Integrals: Class 12 Maths Chapter 7 Hinglish mein

Derivative ko ulta chalana, aur area naapna

Mathematics Part 2 · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

165 min
58 practice questions
41 MCQs

Is chapter mein aap kya seekhenge

Standard derivatives ko ulta padh kar antiderivatives nikaalo, aur ek di hui value se constant of integration fix karo
Linearity, substitution aur trigonometric identities use karke integrand ko standard form mein lao
Special forms 1/(x² ± a²), 1/(a² − x²), 1/√(a² − x²), 1/√(x² ± a²) ko integrate karo, aur un quadratic types ko bhi jo completing the square se inhi forms mein badal jaate hain
Rational function ko partial fractions mein todo aur integrate karo; agar woh improper ho to pehle divide karo
Products ko by parts integrate karo, eˣ[f(x) + f′(x)] pattern pehchano, aur √(a² − x²) aur √(x² ± a²) integrate karo
Definite integral ko area ke roop mein samjhao, aur area function aur dono fundamental theorems use karke definite integrals evaluate karo
Badli hui limits wale substitution se aur properties P0–P7 se definite integrals evaluate karo, jinmein even aur odd functions aur a + b − x reflection bhi shaamil hain

NCERT Mathematics Class 12, Chapter 7, pp. 225–291 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

7.1 Introduction

Speedometer batata hai ki bus har pal kitni tez chal rahi hai, lekin yeh nahi batata ki woh kitni door ja chuki hai. Agar tumhe trip ke har moment ki speed pata ho, to kya tum distance nikaal sakte ho? Differentiation distance ko speed mein badalta hai. Is chapter mein hum ulti direction mein jaana seekhenge.

Integral calculus do tarah ke sawaalon ka jawab deta hai. Pehla sawaal differentiation ko ulta chalata hai: ek function diya hai $f,$ to aisa function $F$ dhoondho jiske liye $F^{'} = f .$ Aise $F$ ko antiderivative kehte hain, yaani woh function jiska derivative hai $f .$ Ek hi formula saare antiderivatives ko ek saath cover karta hai; use hum indefinite integral kehte hain, aur is khoj ko hi integration (समाकलन) kehte hain. Doosra sawaal size ka hai: kisi interval par graph aur $x$ -axis ke beech kitna area hai? Woh number definite integral hai.

3/3Problem 2: $y = f (x)$ aur $x$ -axis ke beech ka area nikaalo, jiski limits hain $a$ aur $b .$

Fig. 7.1Woh do problems jinka jawab integral calculus deta hai.

Do panels aamne-saamne. Pehle step mein left panel par sirf paanch chhote tangent segments hain, barabar doori wali x positions par, har ek wahan ke known slope jitna jhuka hua; right panel feeka hai. Doosre step mein left panel mein ek smooth curve banta hai jo har slope mark ko touch karta hai: yahi woh function hai jise hum dhoondh rahe the. Teesre step mein right panel chamak uthta hai: x-axis ke upar ek badhta hua curve, aur do marked points a aur b ke beech uske neeche ka region shaded hai.

Pehli nazar mein in dono ka aapas mein koi lena-dena nahi lagta. Section 7.8 mein hum dekhenge ki dono ka jawab ek hi hai: areas antiderivatives se nikaale ja sakte hain. Yeh result, Fundamental Theorem of Calculus, lambi limit calculations ki jagah sirf ek subtraction rakh deta hai, aur isi wajah se integrals roz ke tools hain, badalte force se kiye gaye work se lekar is chance tak ki koi measurement ek range mein aaye. Yeh chapter pehle antiderivatives nikaalne ka toolkit banata hai, phir definite integrals aur unki un properties par aata hai jo unhe evaluate karna aasaan banati hain.

7.2 Integration: Differentiation ka Inverse Process

Differentiation ek function leta hai aur uska derivative wapas deta hai. Integration is machine ko ulta chalata hai: derivative se shuru karo aur woh function wapas pao jo use paida karta hai. Teen facts jo tum pehle se jaante ho, ulta padhe ja sakte hain:

\begin{aligned} \frac{d}{d x} (\tan x) & = \sec^{2} x \\ \frac{d}{d x} (\frac{x^{4}}{4}) & = x^{3} \\ \frac{d}{d x} (- \cos x) & = \sin x \end{aligned}

To yeh antiderivatives mil gaye: $\tan x$ jiska derivative hai $\sec^{2} x,$ $\frac{x^{4}}{4}$ jiska derivative hai $x^{3},$ aur $- \cos x$ jiska derivative hai $\sin x .$ Lekin sirf yahi nahi hain. Constant ka derivative zero hota hai, isliye $\frac{x^{4}}{4} + 7$ aur $\frac{x^{4}}{4} - \sqrt{2}$ ka derivative bhi hai $x^{3} .$ Kisi antiderivative mein koi bhi real number $C$ jodne se ek aur antiderivative milta hai. Is $C$ ko arbitrary constant ya constant of integration kehte hain; ise saare real numbers par chalane se answers ki poori family ban jaati hai.

Definition

Antiderivative (primitive)

Maan lo $f$ ek interval par defined hai, jise hum kehte hain $I .$ Ek function $F$ ko $I$ par $f$ ka antiderivative (ya primitive) kehte hain agar $F^{'} (x) = f (x)$ ho, har us $x$ ke liye jo is interval mein hai: $I .$ Tab $F + C$ bhi ek antiderivative hai, har real constant ke liye: $C .$

Example $F (x) = x^{2} - 5 x$ ek antiderivative hai $f (x) = 2 x - 5$ ka, poori real line par, yaani $𝐑,$ aur ek aur antiderivative hai $x^{2} - 5 x + 100.$

Graph mein dekho to family $F (x) + C$ ek hi curve hai jo upar-neeche khiskaya gaya hai. Kisi fixed $x$ par har member ka slope same hota hai, yaani $f (x),$ isliye wahan unke tangents parallel hote hain (Fig. 7.2). Ek extra jaankari, jaise koi point jisse curve ko guzarna hi hai, family mein se ek member chun leti hai.

2/3 $x = \frac{π}{3}$ par har member ka slope hai $\cos \frac{π}{3} = \frac{1}{2},$ isliye tangents parallel hain.

Fig. 7.2Antiderivatives ki ek family, jinka derivative hai

f (x) = \cos x .

x ki value 0 se 2 pi tak wala ek grid. Pehle step mein y barabar sin x plus C type ke teen curves bane hain, C barabar minus 3, 0 aur 3 ke liye; har ek same wave ka vertical shift hai aur har ek par uski C value likhi hai. Doosre step mein x barabar pi by 3 par ek vertical dashed line teeno curves ko kaatti hai, aur har crossing par ek chhota tangent segment dikhata hai ki saare tangents parallel hain, aur slope one half hai. Teesre step mein teeno curves feeke ho jaate hain aur ek naya member, y barabar sin x plus 1, marked point (pi by 2, 2) se guzarta hua banta hai.

TheoremAntiderivatives mein sirf ek constant ka farq hota hai

Agar ek interval hai $I,$ aur uske har point par $g$ aur $h$ ka derivative same hai, to $g (x) - h (x)$ constant rehta hai poore $I .$

Maan lo $ϕ (x) = g (x) - h (x) .$ Tab $ϕ^{'} (x) = g^{'} (x) - h^{'} (x) = 0$ hai, har us $x$ ke liye jo is interval mein hai: $I .$ Ab koi do points lo $x_{1} < x_{2}$ , dono is interval se: $I .$ Mean value theorem se unke beech ek $c$ hai jiske liye $ϕ (x_{2}) - ϕ (x_{1}) = ϕ^{'} (c) (x_{2} - x_{1}) = 0.$ Yaani $ϕ$ ki value same rehti hai poore $I .$

Yahan interval shabd zaroori hai. Maan lo set hai $x \neq 0,$ jiske do tukde hain. Us par woh function jo $x > 0$ ke liye $\log ∣ x ∣$ hai aur $x < 0$ ke liye $\log ∣ x ∣ + 5$ hai, uska derivative bhi hai $\frac{1}{x},$ phir bhi woh $\log ∣ x ∣$ plus ek constant nahi hai. Lekin kisi ek single interval ke andar family ${F + C}$ har antiderivative ko pakad leti hai.

Isi wajah se ek symbol poori family ke liye kaam kar sakta hai. Hum likhte hain $\int f (x) d x,$ aur padhte hain “ $x$ ke respect mein $f$ ka integral”, aur agar $F$ koi bhi ek antiderivative hai, to

Indefinite integral

\begin{array}{c} \int f (x) d x \\ = F (x) + C \end{array}

Isi tarah, agar $\frac{d y}{d x} = f (x)$ ho to hum likhte hain $y = \int f (x) d x .$ Is symbol ke saath aane wale words Table 7.1 mein jama kiye gaye hain.

Table 7.1Indefinite integrals ki bhashaNCERT Table 7.1 · p. 227

Symbol ya word	Iska matlab
$\int f (x) d x$	antiderivatives ki poori family jinka derivative hai $f,$ aur jinka variable hai $x$
$f (x)$	integrand, yaani woh function jise integrate kiya ja raha hai
$d x$	batata hai ki $x$ integration ka variable hai
integrate	indefinite integral nikaalna
$f$ ka ek integral	koi bhi ek function $F$ jiske liye $F^{'} (x) = f (x)$
integration	differentiation ka ulta: integrals nikaalna
$C$	constant of integration, koi bhi real number

Definition

Indefinite integral

Indefinite integral $\int f (x) d x$ kisi interval par $f$ ke saare antiderivatives ki family hai, jise $F (x) + C$ likhte hain, jahan $F$ koi bhi ek antiderivative hai aur $C$ ek arbitrary constant hai.

Tumhare differentiation toolkit ka har derivative formula right se left padhne par integral formula ban jaata hai. Table 7.2 mein woh formulas hain jo hum baar-baar use karte hain. Har row ko uski right-hand side differentiate karke check kar sakte ho.

Table 7.2Standard integrals: right-hand entry ko differentiate karne se left-hand entry milti haiNCERT Standard formulae · p. 228

$f (x)$	$\int f (x) d x$
$x^{n},$ $n \neq - 1$	$\frac{x^{n + 1}}{n + 1} + C$
$1$	$x + C$
$\frac{1}{x}$	$\log ∣ x ∣ + C$
$e^{x}$	$e^{x} + C$
$a^{x}$	$\frac{a^{x}}{\log a} + C$
$\cos x$	$\sin x + C$
$\sin x$	$- \cos x + C$
$\sec^{2} x$	$\tan x + C$
${c o s e c}^{2} x$	$- \cot x + C$
$\sec x \tan x$	$\sec x + C$
$c o s e c x \cot x$	$- c o s e c x + C$
$\frac{1}{\sqrt{1 - x^{2}}}$	$\sin^{- 1} x + C$ ya $- \cos^{- 1} x + C$
$\frac{1}{1 + x^{2}}$	$\tan^{- 1} x + C$ ya $- \cot^{- 1} x + C$

Aakhri do rows mein do-do answers hain. Yeh ek doosre ko contradict nahi karte: $\sin^{- 1} x$ aur $- \cos^{- 1} x$ mein bas ek constant ka farq hai, $\frac{π}{2},$ isliye dono sahi hain. Har formula sirf wahin valid hai jahan dono sides defined hon. $\int \frac{1}{x} d x = \log ∣ x ∣ + C$ har us interval par valid hai jismein yeh point na ho: $0,$ aur square root wale formula mein root ke andar ki quantity positive honi chahiye. Yeh intervals kam hi likhe jaate hain, isliye inhe khud check karne ki aadat daalo.

7.2.1 Indefinite integral ki kuch properties

Neeche ke rules se hum ek complicated integrand ko aise tukdon mein tod sakte hain jinhe Table 7.2 se seedha padh sakein.

PropertyIntegration aur differentiation ek doosre ko undo karte hain

(I) Indefinite integral ko differentiate karne se integrand wapas milta hai, $\frac{d}{d x} \int u d x = u;$ derivative ko integrate karne se function wapas milta hai, ek constant tak, $\int u^{'} d x = u + C .$ (II) Agar do indefinite integrals ka derivative same hai, to woh curves ki ek hi family describe karte hain; tab hum unhe barabar likhte hain.

(I) Agar $\int u d x = U + C$ hai aur $U^{'} = u,$ to $\frac{d}{d x} (U + C) = U^{'} = u .$ Doosre hisse ke liye: $u$ khud ek function hai jiska derivative hai $u^{'},$ isliye yeh is derivative ka ek antiderivative hai: $u^{'},$ aur family hai $u + C .$

(II) Maan lo $\frac{d}{d x} \int f d x = \frac{d}{d x} \int g d x .$ Tab dono integrals ke difference ka derivative interval par $0$ hai, isliye upar wale theorem se woh ek constant hai. Har family doosri family ko ek constant se shift karke milti hai, yaani dono ek hi family hain. $\int f d x = \int g d x$ jaisi equation ko isi sense mein padha jaata hai: arbitrary constant tak barabar.

PropertyLinearity

(III) Sum ka integral integrals ka sum hota hai: $\int (u + v) d x = \int u d x + \int v d x .$ (IV) Constant factor bahar aa sakta hai: $\int k u d x = k \int u d x$ , jahan koi bhi real number ho sakta hai: $k .$ (V) Dono ko milakar: maan lo functions hain $u_{1}, \dots, u_{n}$ aur constants hain $k_{1}, \dots, k_{n},$ to $k_{1} u_{1} + \dots + k_{n} u_{n}$ ka integral hai $k_{1} \int u_{1} d x + \dots + k_{n} \int u_{n} d x .$

(III) ki right-hand side ko term by term differentiate karo: (I) se har integral apna integrand wapas deta hai, isliye derivative hai $u + v .$ (I) se hi left-hand side ka derivative bhi hai $u + v,$ aur same derivative wali do families (II) se barabar hain.

(IV) ke liye: $k \int u d x$ ka derivative hai $k u,$ jo is integral ka bhi derivative hai: $\int k u d x;$ phir se (II) lagao. (III) aur (IV) ko baar-baar use karne se (V) milta hai.

In rules ke saath integrate karne ka sabse simple tareeka hai method of inspection: aisa function guess karo jiska derivative integrand ho, phir apne guess ko differentiate karke confirm karo. Agar check mein constant factor ka farq aaye to use adjust kar lo.

Example 1NCERT Example 1 · p. 231

Inspection se antiderivatives

Har function ka ek antiderivative likho: (i) $\sin 3 x$ (ii) $5 x^{4} - 6 x^{2}$ (iii) $\frac{3}{x},$ $x \neq 0.$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 2NCERT Example 2 · p. 232

Linearity ka use

Nikaalo: (i) $\int \frac{x^{4} + 2}{x^{2}} d x$ (ii) $\int (2 x^{3 / 4} - 5) d x$ (iii) $\int (x^{1 / 3} + 3 e^{x} - \frac{4}{x}) d x .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 3NCERT Example 3 · p. 233

Trigonometric integrands

Nikaalo: (i) $\int (3 \cos x - 2 \sin x) d x$ (ii) $\int \sec x (\sec x - \tan x) d x$ (iii) $\int \frac{1 + \cos x}{\sin^{2} x} d x .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 4NCERT Example 4 · p. 233

Velocity se position tak

Ek toy train seedhe track par chalti hai. $t$ seconds baad uski velocity $v (t) = 3 t^{2} + 2 t$ hai ( $c m / s$ mein), aur $t = 0$ par woh track ke start se $5 c m$ door hai. Uski position $s (t),$ nikaalo, aur yeh bhi ki woh kahan hai jab $t = 2.$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 4 general pattern dikhata hai. Family $F + C$ mein infinitely many antiderivatives hain; ek extra condition, jaise starting value, unmein se exactly ek ko chun leti hai. Variable ka $x$ hona zaroori nahi: $\int t^{6} d t = \frac{t^{7}}{7} + C$ aur $\int u^{- 3} d u = - \frac{1}{2 u^{2}} + C$ bhi same rules follow karte hain.

Practice · 5 questions

Practice 7.1: answers ke saath

1
Short answerEasyNCERT Exercise 7.1, Q1–Q5
Har function ka ek antiderivative guess karo, phir differentiate karke use check karo.
1. $\cos 4 x$
  Solution dikhao
  Answer $\frac{1}{4} \sin 4 x$
  $\frac{d}{d x} \sin 4 x = 4 \cos 4 x,$ isliye $4$ se divide karo: $\frac{d}{d x} (\frac{1}{4} \sin 4 x) = \cos 4 x .$
2. $\sin 5 x$
  Solution dikhao
  Answer $- \frac{1}{5} \cos 5 x$
  $\frac{d}{d x} \cos 5 x = - 5 \sin 5 x,$ isliye $- \frac{1}{5} \cos 5 x$ ka derivative hai $\sin 5 x .$
2
Short answerEasyNCERT Exercise 7.1, Q6–Q13
Har integral nikaalo.
1. $\int (2 e^{2 x} - 3) d x$
  Solution dikhao
  Answer $e^{2 x} - 3 x + C$
  $\int 2 e^{2 x} d x = e^{2 x}$ (check: derivative $2 e^{2 x}$ ) aur $\int 3 d x = 3 x .$
2. $\int x^{3} (1 + \frac{2}{x^{2}}) d x$
  Solution dikhao
  Answer $\frac{x^{4}}{4} + x^{2} + C$
  Expand karo: $x^{3} + 2 x .$ Phir $\int (x^{3} + 2 x) d x = \frac{x^{4}}{4} + x^{2} + C .$
3
Short answerMediumNCERT Exercise 7.1, Q14–Q20
Har integral nikaalo.
1. $\int (2 + x) \sqrt{x} d x$
  Solution dikhao
  Answer $\frac{4}{3} x^{3 / 2} + \frac{2}{5} x^{5 / 2} + C$
  Expand karo: $2 x^{1 / 2} + x^{3 / 2},$ jisse milta hai $2 \cdot \frac{2}{3} x^{3 / 2} + \frac{2}{5} x^{5 / 2} + C .$
2. $\int \sqrt{x} (2 x^{2} - x + 4) d x$
  Solution dikhao
  Answer $\frac{4}{7} x^{7 / 2} - \frac{2}{5} x^{5 / 2} + \frac{8}{3} x^{3 / 2} + C$
  Expand karo: $2 x^{5 / 2} - x^{3 / 2} + 4 x^{1 / 2} .$
  Integrate karo: $2 \cdot \frac{2}{7} x^{7 / 2} - \frac{2}{5} x^{5 / 2} + 4 \cdot \frac{2}{3} x^{3 / 2} + C .$
Quick answerEasyNCERT Exercise 7.1, Q21
$2 \sqrt{x} - \frac{3}{\sqrt{x}}$ ka ek antiderivative hai
$\frac{4}{3} x^{3 / 2} - \frac{3}{2} x^{1 / 2} + C$ $x^{- 1 / 2} + \frac{3}{2} x^{- 3 / 2} + C$ $\frac{4}{3} x^{3 / 2} - 6 x^{1 / 2} + C$ $3 x^{3 / 2} - 6 x^{1 / 2} + C$
Answer dekho
Answer (C) $\frac{4}{3} x^{3 / 2} - 6 x^{1 / 2} + C$
$\int 2 x^{1 / 2} d x = \frac{4}{3} x^{3 / 2}$ aur $\int 3 x^{- 1 / 2} d x = 6 x^{1 / 2} .$ Option B integrate karne ki jagah differentiate kar deta hai; option A exponent se divide karne ki jagah usse multiply kar deta hai.
Quick answerMediumNCERT Exercise 7.1, Q22
Agar $\frac{d}{d x} f (x) = 6 x^{2} - \frac{2}{x^{3}}$ aur $f (1) = 4,$ to $f (x)$ hai
$2 x^{3} - \frac{1}{x^{2}} + 3$ $2 x^{3} + \frac{1}{x^{2}} + 1$ $2 x^{3} + \frac{1}{x^{2}} + 4$ $12 x + \frac{6}{x^{4}} - 14$
Answer dekho
Answer (B) $2 x^{3} + \frac{1}{x^{2}} + 1$
$\int (6 x^{2} - 2 x^{- 3}) d x = 2 x^{3} + x^{- 2} + C,$ aur $f (1) = 3 + C = 4$ se milta hai $C = 1.$ Option A mein $\int - 2 x^{- 3} d x$ ka sign galat hai; option C bhool jaata hai ki $f (1)$ mein pehle se hi yeh shaamil hai: $2 + 1.$

7.3 Integration ke Methods

Inspection tabhi kaam karta hai jab hum integrand ko kisi derivative ke roop mein pehchaan lein, aur zyadatar integrands itne seedhe nahi hote. Isliye hume aise tareeke chahiye jo integral ko tab tak badlein jab tak woh Table 7.2 ki kisi row se match na kar jaaye. Teen tareeke sabse zaroori hain, aur har ek differentiation ke kisi rule se aata hai: substitution chain rule ko undo karta hai, partial fractions fractions jodne ko undo karte hain, aur integration by parts product rule ko undo karta hai.

7.3.1 Integration by substitution

Chain rule kehta hai ki agar $u = g (x)$ ho to $\frac{d}{d x} F (g (x)) = F^{'} (g (x)) g^{'} (x) .$ Ise ulta padho: ( $g (x)$ ka function) × $g^{'} (x)$ type ka integrand kisi simple cheez ka derivative hai. Fig. 7.3 dikhata hai ki ise kaise pehchaanein.

2/3 $\int \cos (x^{2}) 2 x d x$ mein aisa inner function dhoondho jiska derivative bhi wahan maujood ho.

Fig. 7.3Substitution chain rule ko ulta padhna hai.

Pehle step mein teen boxes ki pipeline hai: x, phir u barabar x squared, phir sin u, aur differentiate likhe arrow se cos of x squared times 2x milta hai. Doosre step mein integrand cos of x squared times 2x dikhaya gaya hai; cos of x squared ke around ek solid box hai jis par function of x squared likha hai, aur factor 2x ke around ek dashed box hai jis par derivative of x squared likha hai. Teesre step mein dashed factor aur dx milkar du ban jaate hain, integral ban jaata hai integral of cos u du, aur result sin of x squared plus C dikhta hai.

Symbols mein: agar hum likhein $x = ϕ (u),$ to $d x = ϕ^{'} (u) d u,$ aur poore integral ko $u$ ke terms mein dobara likha ja sakta hai:

Variable badalna

\begin{array}{c} \int f (x) d x \\ = \int f (ϕ (u)) ϕ^{'} (u) d u \end{array}

Practice mein hum zyadatar ulta chalte hain aur integrand ke ek hisse ko naam dete hain: $t = h (x),$ $d t = h^{'} (x) d x .$ $t$ chunna hi woh skill hai jo develop karni hai. Integrand ka aisa hissa dhoondho jiska derivative bhi maujood ho, shayad kisi constant factor ke saath; wahi hissa natural choice hai.

Step by step

Substitution ka working rule

Step 1 / 4

Chuno

Ek inner expression $t = h (x)$ chuno jiska derivative $h^{'} (x)$ integrand mein maujood ho (shayad kisi constant ke saath).

1 / 4

Example 5NCERT Example 5 · p. 236

Chaar substitutions

$x$ ke respect mein integrate karo: (i) $\cos (5 x - 2)$ (ii) $3 x^{2} e^{x^{3}}$ (iii) $\frac{(\log x)^{3}}{x},$ $x > 0$ (iv) $\frac{e^{\sin^{- 1} x}}{\sqrt{1 - x^{2}}} .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

(i) mein inner function linear tha. Yeh case alag se yaad rakhne layak hai: agar $\int f (x) d x = F (x) + C,$ to $\int f (a x + b) d x = \frac{1}{a} F (a x + b) + C$ hota hai, bas shart hai $a \neq 0.$

Substitution un chaar trigonometric functions ko bhi handle karta hai jo hamari table mein nahi hain. Neeche ke results yaad kar lo: yeh baar-baar aate hain.

Derivationtan, cot, sec aur cosec ke integrals

tan. $\tan x = \frac{\sin x}{\cos x} .$ Rakho $t = \cos x,$ $d t = - \sin x d x$ :

\begin{aligned} \int \tan x d x \\ = - \int \frac{d t}{t} \\ = - \log ∣ \cos x ∣ \end{aligned}

plus ek constant. Kyunki $- \log ∣ \cos x ∣ = \log ∣ \sec x ∣,$ isliye yeh hai $\int \tan x d x = \log ∣ \sec x ∣ + C .$

cot. $\cot x = \frac{\cos x}{\sin x};$ rakho $t = \sin x,$ $d t = \cos x d x,$ to milta hai $\int \cot x d x = \log ∣ \sin x ∣ + C .$

sec. Upar aur neeche dono ko multiply karo is se: $\sec x + \tan x .$ Naya numerator $\sec^{2} x + \sec x \tan x$ bilkul naye denominator ka derivative hai, isliye $t = \sec x + \tan x$ lene par:

\begin{aligned} \int \sec x d x \\ = \int \frac{d t}{t} \\ = \log ∣ t ∣ + C \end{aligned}

Yaani $\int \sec x d x = \log ∣ \sec x + \tan x ∣ + C .$

cosec. Upar aur neeche dono ko multiply karo is se: $c o s e c x - \cot x .$ Numerator ban jaata hai ${c o s e c}^{2} x - c o s e c x \cot x,$ jo $t = c o s e c x - \cot x$ ka derivative hai (check: $\frac{d}{d x} c o s e c x = - c o s e c x \cot x$ aur $\frac{d}{d x} (- \cot x) = {c o s e c}^{2} x$ ). To phir se:

\begin{aligned} \int c o s e c x d x \\ = \int \frac{d t}{t} \\ = \log ∣ t ∣ + C \end{aligned}

Yaani $\int c o s e c x d x = \log ∣ c o s e c x - \cot x ∣ + C .$

Kyunki $(c o s e c x - \cot x) (c o s e c x + \cot x) = 1,$ isliye yahi answer aise bhi likh sakte hain: $- \log ∣ c o s e c x + \cot x ∣ + C .$

Example 6NCERT Example 6 · p. 238

Thodi taiyaari ke saath substitution

Nikaalo: (i) $\int \sin^{3} x \cos^{4} x d x$ (ii) $\int \frac{\cos x}{\cos (x - b)} d x$ (iii) $\int \frac{\sin x}{\sin x - \cos x} d x .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Practice · 7 questions

Practice 7.2: answers ke saath

1
Short answerEasyNCERT Exercise 7.2, Q1–Q8
Har function ko kisi suitable substitution se integrate karo.
1. $\frac{3 x^{2}}{4 + x^{3}}$
  Solution dikhao
  Answer $\log ∣ 4 + x^{3} ∣ + C$
  $t = 4 + x^{3},$ $d t = 3 x^{2} d x$ : $\int \frac{d t}{t} = \log ∣ t ∣ + C .$
2. $\frac{(\log x)^{4}}{x}$
  Solution dikhao
  Answer $\frac{(\log x)^{5}}{5} + C$
  $t = \log x,$ $d t = \frac{d x}{x}$ : $\int t^{4} d t = \frac{t^{5}}{5} .$
2
Short answerMediumNCERT Exercise 7.2, Q9–Q16
Har function integrate karo.
1. $(6 x - 3) \sqrt{x^{2} - x + 2}$
  Solution dikhao
  Answer $2 (x^{2} - x + 2)^{3 / 2} + C$
  $t = x^{2} - x + 2,$ $d t = (2 x - 1) d x,$ aur $6 x - 3 = 3 (2 x - 1)$ : $3 \int t^{1 / 2} d t = 2 t^{3 / 2} + C .$
2. $\frac{1}{x + \sqrt{x}},$ $x > 0$
  Solution dikhao
  Answer $2 \log (\sqrt{x} + 1) + C$
  Factorise karo: $\frac{1}{\sqrt{x} (\sqrt{x} + 1)} .$ Rakho $t = \sqrt{x} + 1,$ $d t = \frac{d x}{2 \sqrt{x}}$ : $2 \int \frac{d t}{t} = 2 \log t + C .$
3
Short answerMediumNCERT Exercise 7.2, Q17–Q24
Har function integrate karo.
1. $x^{2} e^{- x^{3}}$
  Solution dikhao
  Answer $- \frac{1}{3} e^{- x^{3}} + C$
  $t = - x^{3},$ $d t = - 3 x^{2} d x$ : $- \frac{1}{3} \int e^{t} d t .$
2. $\frac{(\tan^{- 1} x)^{2}}{1 + x^{2}}$
  Solution dikhao
  Answer $\frac{1}{3} (\tan^{- 1} x)^{3} + C$
  $t = \tan^{- 1} x,$ $d t = \frac{d x}{1 + x^{2}}$ : $\int t^{2} d t = \frac{t^{3}}{3} .$
4
Short answerMediumNCERT Exercise 7.2, Q25–Q32
Har function integrate karo.
1. $\frac{\sec^{2} x}{(2 + \tan x)^{3}}$
  Solution dikhao
  Answer $- \frac{1}{2 (2 + \tan x)^{2}} + C$
  $t = 2 + \tan x,$ $d t = \sec^{2} x d x$ : $\int t^{- 3} d t = - \frac{1}{2 t^{2}} .$
2. $\frac{\sin \sqrt{x}}{\sqrt{x}}$
  Solution dikhao
  Answer $- 2 \cos \sqrt{x} + C$
  $t = \sqrt{x},$ $d t = \frac{d x}{2 \sqrt{x}}$ : $2 \int \sin t d t = - 2 \cos t .$
5
Short answerHardNCERT Exercise 7.2, Q33–Q37
Har function integrate karo.
1. $\frac{\cos x}{\cos x + 3 \sin x}$
  Solution dikhao
  Answer $\frac{x}{10} + \frac{3}{10} \log ∣ \cos x + 3 \sin x ∣ + C$
  Likho $\cos x = A (\cos x + 3 \sin x) + B (3 \cos x - \sin x),$ jahan doosra bracket denominator ka derivative hai.
  $\cos x$ ko match karne par: $A + 3 B = 1;$ $\sin x$ ko match karne par: $3 A - B = 0.$ To $A = \frac{1}{10},$ $B = \frac{3}{10},$ aur integral hai $A x + B \log ∣ \cos x + 3 \sin x ∣ + C .$
2. $\frac{\sqrt{\cot x}}{\sin x \cos x},$ $0 < x < \frac{π}{2}$
  Solution dikhao
  Answer $- 2 \sqrt{\cot x} + C$
  $\frac{1}{\sin x \cos x} = \frac{{c o s e c}^{2} x}{\cot x},$ isliye integrand hai $(\cot x)^{- 1 / 2} {c o s e c}^{2} x .$
  $t = \cot x,$ $d t = - {c o s e c}^{2} x d x$ : $- \int t^{- 1 / 2} d t = - 2 \sqrt{t} .$
Quick answerEasyNCERT Exercise 7.2, Q38
$\int \frac{3 x^{2} + 2^{x} \log 2}{x^{3} + 2^{x}} d x$ barabar hai
$(x^{3} + 2^{x})^{- 1} + C$ $\log ∣ 3 x^{2} + 2^{x} \log 2 ∣ + C$ $x^{3} + 2^{x} + C$ $\log ∣ x^{3} + 2^{x} ∣ + C$
Answer dekho
Answer (D) $\log ∣ x^{3} + 2^{x} ∣ + C$
Numerator bilkul derivative hai is expression ka: $x^{3} + 2^{x},$ isliye integral hai $\log ∣ x^{3} + 2^{x} ∣ + C .$ Option B galat expression ka log leta hai.
Quick answerMediumNCERT Exercise 7.2, Q39
$\int \frac{4 d x}{\sin^{2} 2 x}$ barabar hai
$- \cot 2 x + C$ $\tan x - \cot x + C$ $4 \cot 2 x + C$ $\tan x + \cot x + C$
Answer dekho
Answer (B) $\tan x - \cot x + C$
$\frac{4}{\sin^{2} 2 x} = \frac{1}{\sin^{2} x \cos^{2} x} = \frac{\sin^{2} x + \cos^{2} x}{\sin^{2} x \cos^{2} x} = \sec^{2} x + {c o s e c}^{2} x,$ jiska integral hai $\tan x - \cot x .$ (Yeh barabar hai $- 2 \cot 2 x,$ isliye option A mein 2 ke factor ka farq hai.)

7.3.2 Trigonometric identities se integration

Sines aur cosines ke squares, cubes aur products hamari table mein nahi hain, lekin identities unhe single sines aur cosines ke sum mein likh deti hain, jo table mein hain. Teen sabse kaam ke formulas hain: double-angle formulas is form mein $\cos^{2} x = \frac{1 + \cos 2 x}{2}$ aur $\sin^{2} x = \frac{1 - \cos 2 x}{2},$ product-to-sum formulas jaise $2 \cos A \cos B = \cos (A + B) + \cos (A - B),$ aur triple-angle formulas. Fig. 7.4 dikhata hai ki pehla formula sach kyun hai.

3/3Yeh line $y = \frac{1}{2}$ ke around $\frac{1}{2}$ height ki wave hai: $\cos^{2} x = \frac{1}{2} + \frac{1}{2} \cos 2 x .$

Fig. 7.4Cosine ko square karne se ek constant ke upar ek tez, chhota cosine milta hai.

x ki value 0 se 2 pi tak ka ek graph. Pehle step mein cosine curve minus 1 aur 1 ke beech jhoolta hai. Doosre step mein curve cos squared x joda jaata hai: yeh kabhi 0 se neeche nahi jaata, 0, pi aur 2 pi par 1 ko touch karta hai, pi by 2 aur 3 pi by 2 par 0 ko touch karta hai, aur do guna jaldi repeat hota hai. Teesre step mein one half height par ek dashed horizontal line aati hai aur dikhta hai ki cos squared curve us line par savaar ek cosine wave hai jiska amplitude one half aur period pi hai; yahi identity hai cos squared x barabar one half plus one half cos 2x.

Example 7NCERT Example 7 · p. 241

Teen identities kaam karte hue

Nikaalo: (i) $\int \sin^{2} 3 x d x$ (ii) $\int \cos 5 x \cos 2 x d x$ (iii) $\int \cos^{3} x d x .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Kaun si identity

\sin^{2} 4 x

ko aisi cheez mein badalti hai jise hum seedha integrate kar sakein?

\sin^{2} 4 x = 1 - \cos^{2} 4 x

\sin^{2} 4 x = 2 \sin 2 x \cos 2 x

\sin^{2} 4 x = \frac{1 - \cos 4 x}{2}

\sin^{2} 4 x = \frac{1 - \cos 8 x}{2}

Answer dekho

Answer (D) $\sin^{2} 4 x = \frac{1 - \cos 8 x}{2}$

Double-angle form square hata deta hai:

\frac{1}{2} - \frac{1}{2} \cos 8 x

term by term integrate ho jaata hai. Option C galat angle use karta hai (use

4 x

ko double karke

8 x

banana chahiye tha); option A bas ek square ki jagah doosra square rakh deta hai.

Practice · 5 questions

Practice 7.3: answers ke saath

1
Short answerMediumNCERT Exercise 7.3, Q1–Q8
Har function ko kisi identity se integrate karo.
1. $\cos^{2} (3 x - 1)$
  Solution dikhao
  Answer $\frac{x}{2} + \frac{1}{12} \sin (6 x - 2) + C$
  $\cos^{2} θ = \frac{1 + \cos 2 θ}{2}$ , jahan $θ = 3 x - 1$ : $\int \frac{1 + \cos (6 x - 2)}{2} d x .$
2. $\sin 5 x \cos 2 x$
  Solution dikhao
  Answer $- \frac{1}{14} \cos 7 x - \frac{1}{6} \cos 3 x + C$
  $\sin 5 x \cos 2 x = \frac{1}{2} (\sin 7 x + \sin 3 x) .$
2
Short answerMediumNCERT Exercise 7.3, Q9–Q16
Har function integrate karo.
1. $\frac{\cos x}{1 - \cos x}$
  Solution dikhao
  Answer $- \cot \frac{x}{2} - x + C$
  $\frac{\cos x}{1 - \cos x} = \frac{1}{1 - \cos x} - 1 = \frac{1}{2} {c o s e c}^{2} \frac{x}{2} - 1.$
2. $\cos^{4} x$
  Solution dikhao
  Answer $\frac{3 x}{8} + \frac{\sin 2 x}{4} + \frac{\sin 4 x}{32} + C$
  $\cos^{4} x = (\frac{1 + \cos 2 x}{2})^{2} = \frac{1}{4} (1 + 2 \cos 2 x + \cos^{2} 2 x),$ aur $\cos^{2} 2 x = \frac{1 + \cos 4 x}{2} .$
  To $\cos^{4} x = \frac{3}{8} + \frac{1}{2} \cos 2 x + \frac{1}{8} \cos 4 x .$
3
Short answerHardNCERT Exercise 7.3, Q17–Q22
Har function integrate karo.
1. $\frac{\sin^{3} x - \cos^{3} x}{\sin^{2} x \cos^{2} x}$
  Solution dikhao
  Answer $\sec x + c o s e c x + C$
  Todo: $\frac{\sin x}{\cos^{2} x} - \frac{\cos x}{\sin^{2} x} = \sec x \tan x - c o s e c x \cot x .$
2. $\frac{\cos 2 x - 2 \cos^{2} x}{\sin^{2} x}$
  Solution dikhao
  Answer $\cot x + C$
  $\cos 2 x - 2 \cos^{2} x = - 1,$ isliye integrand hai $- {c o s e c}^{2} x .$
Quick answerMediumNCERT Exercise 7.3, Q23
$\int (\tan x - \cot x)^{2} d x$ barabar hai
$\tan x - \cot x - 4 x + C$ $\tan x + \cot x - 2 x + C$ $\frac{1}{3} (\tan x - \cot x)^{3} + C$ $\sec^{2} x + {c o s e c}^{2} x - 4 x + C$
Answer dekho
Answer (A) $\tan x - \cot x - 4 x + C$
$(\tan x - \cot x)^{2} = \tan^{2} x + \cot^{2} x - 2 = \sec^{2} x + {c o s e c}^{2} x - 4,$ jiska integral hai $\tan x - \cot x - 4 x .$ Option C bracket ko power rule mein $x$ ki tarah treat karta hai, jo galat hai kyunki uska derivative 1 nahi hai.
Quick answerMediumNCERT Exercise 7.3, Q24
$\int (1 + \log x) \sec^{2} (x \log x) d x$ barabar hai
$\sec (x \log x) + C$ $x \tan (\log x) + C$ $\log ∣ \tan x ∣ + C$ $\tan (x \log x) + C$
Answer dekho
Answer (D) $\tan (x \log x) + C$
$\frac{d}{d x} (x \log x) = 1 + \log x,$ isliye $t = x \log x$ lene par integral hai $\int \sec^{2} t d t = \tan t + C .$

7.4 Kuch Khaas Functions ke Integrals

$x^{2}$ aur ek constant $a^{2}$ (jahan $a > 0$ ) wale chhe integrands itni baar aate hain ki hum unhe ek baar prove karke phir seedhe use karte hain. Completing the square ke saath milkar yeh har us integrand ko handle karte hain jo is form ka ho: $\frac{1}{quadratic}$ ya $\frac{1}{\sqrt{quadratic}} .$

Table 7.3Chhe special integrals (inki numbering poore chapter mein use hoti hai)NCERT 7.4 (1)–(6) · p. 243

	Integrand	Integral ( $C$ jodo)
(1)	$\frac{1}{x^{2} - a^{2}}$	$\frac{1}{2 a} \log ∣ \frac{x - a}{x + a} ∣$
(2)	$\frac{1}{a^{2} - x^{2}}$	$\frac{1}{2 a} \log ∣ \frac{a + x}{a - x} ∣$
(3)	$\frac{1}{x^{2} + a^{2}}$	$\frac{1}{a} \tan^{- 1} \frac{x}{a}$
(4)	$\frac{1}{\sqrt{x^{2} - a^{2}}}$	$\log ∣ x + \sqrt{x^{2} - a^{2}} ∣$
(5)	$\frac{1}{\sqrt{a^{2} - x^{2}}}$	$\sin^{- 1} \frac{x}{a}$
(6)	$\frac{1}{\sqrt{x^{2} + a^{2}}}$	$\log ∣ x + \sqrt{x^{2} + a^{2}} ∣$

DerivationYeh chhe formulas sahi kyun hain

(1) aur (2): fraction ko todo. Kyunki $(x + a) - (x - a) = 2 a,$

\begin{aligned} \frac{1}{x^{2} - a^{2}} \\ = \frac{1}{2 a} (\frac{1}{x - a} - \frac{1}{x + a}) \end{aligned}

Integrate karne par milta hai $\frac{1}{2 a} (\log ∣ x - a ∣ - \log ∣ x + a ∣),$ jo (1) hai. Isi tarah $\frac{1}{a^{2} - x^{2}} = \frac{1}{2 a} (\frac{1}{a - x} + \frac{1}{a + x});$ aur kyunki $\int \frac{d x}{a - x} = - \log ∣ a - x ∣,$ humein (2) milta hai. Fraction ko is tarah todna hi Section 7.5 ka topic hai.

(3) aur (5): ek trigonometric substitution. (3) ke liye rakho $x = a \tan θ,$ to $d x = a \sec^{2} θ d θ$ aur $x^{2} + a^{2} = a^{2} \sec^{2} θ$ :

\begin{aligned} \int \frac{a \sec^{2} θ d θ}{a^{2} \sec^{2} θ} = \frac{θ}{a} \\ = \frac{1}{a} \tan^{- 1} \frac{x}{a} \end{aligned}

(5) ke liye rakho $x = a \sin θ;$ tab $\sqrt{a^{2} - x^{2}} = a \cos θ$ aur $d x = a \cos θ d θ,$ isliye integral hai $\int d θ = θ = \sin^{- 1} \frac{x}{a} .$

(4) aur (6): differentiate karke check karo. Maan lo $s = \sqrt{x^{2} \pm a^{2}},$ to $\frac{d s}{d x} = \frac{x}{s} .$ Tab

\begin{aligned} \frac{d}{d x} \log ∣ x + s ∣ \\ = \frac{1 + \frac{x}{s}}{x + s} = \frac{1}{s} \end{aligned}

jo bilkul integrand hai. (Yahi answers in substitutions se bhi aate hain: $x = a \sec θ$ aur $x = a \tan θ,$ saath mein Section 7.3.1 ka $\int \sec θ d θ$ ; us raaste se ek extra term aata hai, $- \log a,$ jo ek constant hai aur $C$ mein mil jaata hai.)

Quadratics general mein. Integrand $\frac{1}{a x^{2} + b x + c}$ ya $\frac{1}{\sqrt{a x^{2} + b x + c}}$ ko completing the square se chhe forms mein se kisi ek mein laaya jaata hai. Bahar nikaalo $a,$ phir $x$ ke coefficient ke aadhe ka square jodo aur ghatao:

Completing the square

\begin{aligned} a x^{2} + b x + c \\ = a [x^{2} + \frac{b}{a} x + \frac{c}{a}] \\ = a [(x + \frac{b}{2 a})^{2} \\ + \frac{c}{a} - \frac{b^{2}}{4 a^{2}}] \end{aligned}

$t = x + \frac{b}{2 a}$ lene par (to $d t = d x$ ) bracket ban jaata hai $t^{2} + k^{2}$ ya $t^{2} - k^{2},$ yeh is baat par depend karta hai ki is expression ka sign kya hai: $\frac{c}{a} - \frac{b^{2}}{4 a^{2}} .$ Fig. 7.5 tiles ke saath yeh idea dikhata hai.

3/4Khaali corner ka area hai $2^{2} = 4,$ isliye $x^{2} + 4 x + 4 = (x + 2)^{2} .$

Fig. 7.5Completing the square, tiles se banaya hua.

Pehle step mein x side ka ek bada square, jis par x squared likha hai, width 4 aur height x wale ek lambe rectangle ke paas rakha hai, jis par 4x likha hai. Doosre step mein rectangle ko width 2 ki do strips mein kaata jaata hai: ek ko ghuma kar square ke neeche le jaate hain, to square ki right side par ek strip aur bottom par ek strip hoti hai, aur ek khaali 2 by 2 corner dashed lines se bana rehta hai. Teesre step mein ek chhota 2 by 2 square corner bhar deta hai, aur x plus 2 side ka square poora ho jaata hai. Chauthe step mein square ke paas likha hai: x squared plus 4x plus 20 barabar x plus 2 ka poora square, plus 16.

Linear over quadratic. $\int \frac{p x + q}{a x^{2} + b x + c} d x$ aur $\int \frac{p x + q}{\sqrt{a x^{2} + b x + c}} d x$ ke liye hum numerator ko quadratic ke derivative ke ek multiple aur ek constant mein todte hain:

Linear numerator

\begin{array}{c} p x + q \\ = A (2 a x + b) + B \end{array}

Step by step

Linear over quadratic ka working rule

Step 1 / 4

Coefficients match karo

$x$ ka coefficient ( $p = 2 a A$ ) aur constant term ( $q = A b + B$ ) compare karke nikaalo $A$ aur $B .$

1 / 4

Example 8NCERT Example 8 · p. 247

Seedha table se

Nikaalo: (i) $\int \frac{d x}{x^{2} - 25}$ (ii) $\int \frac{d x}{\sqrt{6 x - x^{2} - 5}} .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 9NCERT Example 9 · p. 247

Pehle square complete karo

Nikaalo: (i) $\int \frac{d x}{x^{2} + 4 x + 20}$ (ii) $\int \frac{d x}{2 x^{2} + 5 x - 3}$ (iii) $\int \frac{d x}{\sqrt{2 x^{2} + 4 x}} .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 10NCERT Example 10 · p. 249

Linear numerator

Nikaalo: (i) $\int \frac{3 x + 1}{x^{2} + 2 x + 5} d x$ (ii) $\int \frac{x + 1}{\sqrt{8 + 2 x - x^{2}}} d x .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

4 mein se 1

Har integrand ko us formula se match karo jo use poora karta hai.

Left mein koi item tap karo, phir right mein uska match.

Answer dekho

Answer

$\frac{1}{x^{2} + 49}$ : $\frac{1}{7} \tan^{- 1} \frac{x}{7}$
$\frac{1}{\sqrt{49 - x^{2}}}$ : $\sin^{- 1} \frac{x}{7}$
$\frac{1}{x^{2} - 49}$ : $\frac{1}{14} \log ∣ \frac{x - 7}{x + 7} ∣$
$\frac{1}{\sqrt{x^{2} + 49}}$ : $\log ∣ x + \sqrt{x^{2} + 49} ∣$

Plus sign aur root nahi: inverse tangent.

a^{2} - x^{2}

ka root: inverse sine. Minus sign, root nahi: ratio ka logarithm.

x^{2} \pm a^{2}

ka root:

x

plus root ka logarithm.

Practice · 5 questions

Practice 7.4: answers ke saath

1
Short answerEasyNCERT Exercise 7.4, Q1–Q8
Formulas (1)–(6) use karke har function integrate karo.
1. $\frac{4 x^{3}}{x^{8} + 9}$
  Solution dikhao
  Answer $\frac{1}{3} \tan^{- 1} \frac{x^{4}}{3} + C$
  $t = x^{4},$ $d t = 4 x^{3} d x$ : $\int \frac{d t}{t^{2} + 3^{2}},$ formula (3) lagao.
2. $\frac{1}{\sqrt{9 x^{2} + 4}}$
  Solution dikhao
  Answer $\frac{1}{3} \log ∣ 3 x + \sqrt{9 x^{2} + 4} ∣ + C$
  $t = 3 x,$ $d x = \frac{d t}{3}$ : $\frac{1}{3} \int \frac{d t}{\sqrt{t^{2} + 2^{2}}},$ formula (6) lagao.
2
Short answerMediumNCERT Exercise 7.4, Q9–Q15
Square complete karo, phir integrate karo.
1. $\frac{\cos x}{\sqrt{\sin^{2} x + 9}}$
  Solution dikhao
  Answer $\log ∣ \sin x + \sqrt{\sin^{2} x + 9} ∣ + C$
  $t = \sin x$ : $\int \frac{d t}{\sqrt{t^{2} + 3^{2}}},$ formula (6) lagao.
2. $\frac{1}{\sqrt{x^{2} - 6 x + 13}}$
  Solution dikhao
  Answer $\log ∣ x - 3 + \sqrt{x^{2} - 6 x + 13} ∣ + C$
  $x^{2} - 6 x + 13 = (x - 3)^{2} + 2^{2};$ formula (6), jahan $t = x - 3.$
3
Short answerHardNCERT Exercise 7.4, Q16–Q23
Har function integrate karo (linear numerator).
1. $\frac{6 x - 1}{\sqrt{3 x^{2} - x + 2}}$
  Solution dikhao
  Answer $2 \sqrt{3 x^{2} - x + 2} + C$
  Numerator bilkul derivative hai is expression ka: $3 x^{2} - x + 2,$ isliye $t = 3 x^{2} - x + 2$ lene par integral hai $\int t^{- 1 / 2} d t = 2 \sqrt{t} .$
2. $\frac{x - 3}{\sqrt{x^{2} + 4}}$
  Solution dikhao
  Answer $\sqrt{x^{2} + 4} - 3 \log ∣ x + \sqrt{x^{2} + 4} ∣ + C$
  $\int \frac{x d x}{\sqrt{x^{2} + 4}} = \sqrt{x^{2} + 4},$ aur $3 \int \frac{d x}{\sqrt{x^{2} + 4}}$ formula (6) hai.
Quick answerEasyNCERT Exercise 7.4, Q24
$\int \frac{d x}{x^{2} - 4 x + 5}$ barabar hai
$x \tan^{- 1} (x - 2) + C$ $(x - 2) \tan^{- 1} x + C$ $\tan^{- 1} (x - 2) + C$ $\tan^{- 1} x + C$
Answer dekho
Answer (C) $\tan^{- 1} (x - 2) + C$
$x^{2} - 4 x + 5 = (x - 2)^{2} + 1,$ aur $a = 1$ ke saath formula (3) deta hai $\tan^{- 1} (x - 2) .$
Quick answerMediumNCERT Exercise 7.4, Q25
$\int \frac{d x}{\sqrt{5 x - 9 x^{2}}}$ barabar hai
$\frac{1}{9} \sin^{- 1} \frac{18 x - 5}{5} + C$ $\sin^{- 1} \frac{18 x - 5}{5} + C$ $\frac{1}{3} \sin^{- 1} \frac{18 x - 5}{5} + C$ $\frac{1}{3} \sin^{- 1} \frac{9 x - 5}{5} + C$
Answer dekho
Answer (C) $\frac{1}{3} \sin^{- 1} \frac{18 x - 5}{5} + C$
$5 x - 9 x^{2} = 9 [(\frac{5}{18})^{2} - (x - \frac{5}{18})^{2}],$ isliye integral hai $\frac{1}{3} \sin^{- 1} \frac{x - 5 / 18}{5 / 18} = \frac{1}{3} \sin^{- 1} \frac{18 x - 5}{5} .$ Option A $\frac{1}{9}$ use karta hai, jabki hona chahiye $\frac{1}{\sqrt{9}} .$

7.5 Partial Fractions se Integration

Rational function do polynomials ka quotient $\frac{P (x)}{Q (x)}$ hota hai, jahan $Q (x) \neq 0.$ Fractions jodna aasaan hai: $\frac{2}{x - 1} + \frac{1}{x + 2} = \frac{3 x + 3}{(x - 1) (x + 2)} .$ Integration ko iska ulta chahiye. Right wala single fraction diya ho, to hum left wale simple tukde wapas nikaalte hain, jinhe partial fractions (आंशिक भिन्न) kehte hain. Har tukde ka integral ek logarithm, ek power ya ek inverse tangent hota hai.

Definition

Proper aur improper rational functions

$\frac{P (x)}{Q (x)}$ ko proper kehte hain jab upar wale ki degree neeche wale se kam ho ( $\deg P < \deg Q$ ), aur improper jab ho $\deg P \geq \deg Q .$ $P$ ko $Q$ se divide karne par improper fraction ban jaata hai ek polynomial $T (x)$ plus ek proper fraction $\frac{P_{1} (x)}{Q (x)} .$

Example $\frac{x^{2} + 2}{x^{2} - 3 x + 2}$ improper hai (degrees barabar hain) aur barabar hai $1 + \frac{3 x}{x^{2} - 3 x + 2} .$

Division ke baad bacha polynomial part seedha integrate ho jaata hai, isliye asli kaam proper fractions ke saath hai. Table 7.4 har us denominator ko cover karta hai jo humein milega: linear factors ke products, shayad repeated, aur aise quadratics jinke real roots nahi hain. Har factor kuch terms deta hai jinmein unknown constants hote hain, $A, B, C, \dots;$ poore fraction ke partial fractions in sab contributions ka sum hote hain.

Table 7.4Proper fraction ke liye kaun se partial fractions likhne hainNCERT Table 7.2 · p. 253

$Q (x)$ ka factor	Isse judne wale terms
$x - a$	$\frac{A}{x - a}$
$(x - a)^{2}$	$\frac{A}{x - a} + \frac{B}{(x - a)^{2}}$
$x^{2} + b x + c$ , jiske real roots nahi hain	$\frac{B x + C}{x^{2} + b x + c}$

Jaise, $\frac{x + 5}{(x - 1)^{2} (x + 2)}$ ko likhte hain $\frac{A}{x - 1} + \frac{B}{(x - 1)^{2}} + \frac{C}{x + 2},$ aur $\frac{3 x^{2} + 2}{(x - 1) (x^{2} + 4)}$ ko likhte hain $\frac{A}{x - 1} + \frac{B x + C}{x^{2} + 4} .$ Constants nikaalne ke liye poori equation ko $Q (x)$ se multiply karo aur coefficients compare karo, ya phir aisi convenient values rakho jo le sakta hai $x .$ Fig. 7.6 dikhata hai ki todne se graph par kya hota hai.

3/3Dono ko jodne par: $\frac{1}{x - 1} - \frac{1}{x + 1} = \frac{2}{x^{2} - 1},$ jismein dono behaviours hain.

Fig. 7.6Rational function do simple tukdon ke sum ke roop mein.

x ki value minus 4 se 4 tak ka ek coordinate grid. Pehle step mein y barabar 1 over x minus 1 ka graph banta hai, x barabar 1 par ek dashed vertical asymptote ke saath. Doosre step mein y barabar minus 1 over x plus 1 ka graph joda jaata hai, x barabar minus 1 par dashed asymptote ke saath. Teesre step mein dono feeke ho jaate hain aur unka sum, y barabar 2 over x squared minus 1, moti line se banta hai: iske dono asymptotes hain, unke beech x barabar 0 par yeh minus 2 tak neeche jaata hai, aur dono taraf door jaakar 0 ke paas pahunchta hai.

Example 11NCERT Example 11 · p. 254

Do alag linear factors

Nikaalo $\int \frac{d x}{(x - 1) (x + 3)} .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 11 ki line $1 = A (x + 3) + B (x - 1)$ ek identity hai: yeh sach hai chahe jo bhi ho $x,$ sirf kuch values ke liye nahi (kuch books iske liye $\equiv$ likhti hain). Isi se ek tez raasta milta hai: $x = 1$ rakho to milta hai $1 = 4 A,$ aur $x = - 3$ rakho to milta hai $1 = - 4 B .$ Factors ke roots substitute karne se ek baar mein ek ko chhod kar baaki saare unknowns hat jaate hain.

Example 12NCERT Example 12 · p. 254

Improper fraction: pehle divide karo

Nikaalo $\int \frac{x^{2} + 2}{x^{2} - 3 x + 2} d x .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 13NCERT Example 13 · p. 255

Repeated linear factor

Nikaalo $\int \frac{x + 5}{(x - 1)^{2} (x + 2)} d x .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 14NCERT Example 14 · p. 256

x² ko ek variable maano

Nikaalo $\int \frac{x^{2}}{(x^{2} + 4) (x^{2} + 9)} d x .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 14 mein $y$ sirf algebra ke liye ek jugaad tha, integral mein substitution nahi. Agla example pehle ek asli substitution use karta hai, aur uske baad hi partial fractions.

Example 15NCERT Example 15 · p. 256

Pehle substitute, phir todo

Nikaalo $\int \frac{(2 \sin θ + 3) \cos θ}{\cos^{2} θ - 2 \sin θ - 2} d θ .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 16NCERT Example 16 · p. 257

Quadratic factor

Nikaalo $\int \frac{3 x^{2} + 2}{(x - 1) (x^{2} + 4)} d x .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

\frac{2 x - 1}{(x + 3) (x^{2} + 1)}

ke liye shuru karne ka sahi form kaun sa hai?

\frac{A}{x + 3} + \frac{B}{x^{2} + 1}

\frac{A}{x + 3} + \frac{B x + C}{x^{2} + 1}

\frac{A x + B}{x + 3} + \frac{C}{x^{2} + 1}

\frac{A}{x + 3} + \frac{B}{x + 1} + \frac{C}{x - 1}

Answer dekho

Answer (B) $\frac{A}{x + 3} + \frac{B x + C}{x^{2} + 1}$

Irreducible quadratic factor ko linear numerator chahiye:

B x + C .

Option A mein itne unknowns nahi ki har numerator match ho sake, aur option D factorise karta hai

x^{2} + 1,

jiske real roots hain hi nahi.

Practice · 5 questions

Practice 7.5: answers ke saath

1
Short answerEasyNCERT Exercise 7.5, Q1–Q8
Partial fractions mein todo aur integrate karo.
1. $\frac{x}{(x - 2) (x + 3)}$
  Solution dikhao
  Answer $\frac{2}{5} \log ∣ x - 2 ∣ + \frac{3}{5} \log ∣ x + 3 ∣ + C$
  $x = A (x + 3) + B (x - 2)$ : $x = 2$ se $A = \frac{2}{5};$ $x = - 3$ se $B = \frac{3}{5} .$
2. $\frac{3}{x^{2} - 4}$
  Solution dikhao
  Answer $\frac{3}{4} \log ∣ \frac{x - 2}{x + 2} ∣ + C$
  $\frac{3}{(x - 2) (x + 2)} = \frac{3}{4} (\frac{1}{x - 2} - \frac{1}{x + 2}) .$
2
Short answerMediumNCERT Exercise 7.5, Q9–Q16
Partial fractions mein todo aur integrate karo.
1. $\frac{x + 3}{x^{3} + x^{2} - x - 1}$
  Solution dikhao
  Answer $\log ∣ \frac{x - 1}{x + 1} ∣ + \frac{1}{x + 1} + C$
  Factorise karo: $x^{2} (x + 1) - (x + 1) = (x - 1) (x + 1)^{2} .$
  $x + 3 = A (x + 1)^{2} + B (x - 1) (x + 1) + C (x - 1)$ : $x = 1$ se $A = 1;$ $x = - 1$ se $C = - 1;$ $x^{2}$ se $A + B = 0,$ $B = - 1.$
2. $\frac{3 x - 1}{(x^{2} - 1) (x + 2)}$
  Solution dikhao
  Answer $\frac{1}{3} \log ∣ x - 1 ∣ + 2 \log ∣ x + 1 ∣ - \frac{7}{3} \log ∣ x + 2 ∣ + C$
  Teen linear factors. $x = 1$ : $2 = 6 A;$ $x = - 1$ : $- 4 = - 2 B;$ $x = - 2$ : $- 7 = 3 C .$
3
Short answerHardNCERT Exercise 7.5, Q17–Q21
Har function integrate karo.
1. $\frac{\sin x}{(1 + \cos x) (2 + \cos x)}$
  Solution dikhao
  Answer $\log \frac{2 + \cos x}{1 + \cos x} + C$
  $t = \cos x,$ $d t = - \sin x d x$ : $- \int \frac{d t}{(1 + t) (2 + t)} = - \int (\frac{1}{1 + t} - \frac{1}{2 + t}) d t .$
  $= - \log ∣ 1 + t ∣ + \log ∣ 2 + t ∣ .$
2. $\frac{(x^{2} + 2) (x^{2} + 5)}{(x^{2} + 1) (x^{2} + 4)}$
  Solution dikhao
  Answer $x + \frac{4}{3} \tan^{- 1} x + \frac{1}{3} \tan^{- 1} \frac{x}{2} + C$
  $y = x^{2}$ lene par: $\frac{y^{2} + 7 y + 10}{y^{2} + 5 y + 4} = 1 + \frac{2 y + 6}{(y + 1) (y + 4)} .$
  $2 y + 6 = A (y + 4) + B (y + 1)$ : $A = \frac{4}{3},$ $B = \frac{2}{3} .$ To integrand hai $1 + \frac{4 / 3}{x^{2} + 1} + \frac{2 / 3}{x^{2} + 4} .$
Quick answerMediumNCERT Exercise 7.5, Q22
$\int \frac{x d x}{(x + 1) (x + 2)}$ barabar hai
$\log ∣ \frac{(x + 1)^{2}}{x + 2} ∣ + C$ $\log ∣ \frac{(x + 2)^{2}}{x + 1} ∣ + C$ $\log ∣ (x + 1) (x + 2) ∣ + C$ $\log {∣ \frac{x + 2}{x + 1} ∣}^{2} + C$
Answer dekho
Answer (B) $\log ∣ \frac{(x + 2)^{2}}{x + 1} ∣ + C$
$\frac{x}{(x + 1) (x + 2)} = \frac{- 1}{x + 1} + \frac{2}{x + 2},$ isliye integral hai $2 \log ∣ x + 2 ∣ - \log ∣ x + 1 ∣ = \log ∣ \frac{(x + 2)^{2}}{x + 1} ∣ .$ Option A dono factors ke roles ulat deta hai.
Quick answerMediumNCERT Exercise 7.5, Q23
$\int \frac{d x}{x (x^{2} + 4)}$ barabar hai
$\frac{1}{4} \log ∣ x ∣ + \frac{1}{8} \log (x^{2} + 4) + C$ $\log ∣ x ∣ - \frac{1}{2} \log (x^{2} + 4) + C$ $\frac{1}{4} \log ∣ x ∣ - \frac{1}{8} \log (x^{2} + 4) + C$ $\frac{1}{2} \tan^{- 1} \frac{x}{2} + C$
Answer dekho
Answer (C) $\frac{1}{4} \log ∣ x ∣ - \frac{1}{8} \log (x^{2} + 4) + C$
$1 = A (x^{2} + 4) + (B x + C) x$ se milta hai $A = \frac{1}{4},$ $B = - \frac{1}{4},$ $C = 0.$ To integrand hai $\frac{1}{4 x} - \frac{x}{4 (x^{2} + 4)},$ jiska integral hai $\frac{1}{4} \log ∣ x ∣ - \frac{1}{8} \log (x^{2} + 4) .$

7.6 Integration by Parts

Substitution chain rule ko undo karta hai. Integration by parts (खंडशः समाकलन) product rule ko undo karta hai. Yeh products ke liye main tool hai, jaise $x \cos x,$ $x^{2} e^{x}$ ya $e^{x} \sin x,$ aur akele functions ke liye bhi, jaise $\log x$ ya $\tan^{- 1} x .$

DerivationIntegration by parts ka formula

Maan lo $u$ aur $v$ differentiable functions hain, aur dono ka variable hai $x .$ Product rule kehta hai $\frac{d}{d x} (u v) = u \frac{d v}{d x} + v \frac{d u}{d x} .$ Dono sides integrate karo; left side wapas deta hai $u v$ :

\begin{aligned} u v = & \int u \frac{d v}{d x} d x \\ + \int v \frac{d u}{d x} d x \end{aligned}

Rearrange karne par formula apne sabse chhote form mein milta hai, jise aksar likhte hain $\int u d v = u v - \int v d u$ :

\begin{aligned} \int u \frac{d v}{d x} d x \\ = u v - \int v \frac{d u}{d x} d x \end{aligned}

(7.1)

Maan lo product hai $f (x) g (x),$ to $f$ ko first function kaho ( $u = f$ ) aur $g$ ko second function ( $\frac{d v}{d x} = g,$ yaani $v = \int g d x$ ). Tab

Integration by parts

\begin{aligned} \int f g d x \\ = f \int g d x \\ - \int (f^{'} ⁣ \int g d x) d x \end{aligned}

Words mein: first function ko rakho, use second ke integral se multiply karo; phir (first ka derivative) × (second ka integral) ka integral ghatao. Method tab faayda deta hai jab naya integral purane se simple ho. Fig. 7.7 formula ko areas ke baare mein ek statement ki tarah dikhata hai.

4/4Dono milkar banate hain $u_{2} v_{2} - u_{1} v_{1}$ : isliye $\int u d v = [u v] - \int v d u .$

Fig. 7.7Integration by parts do areas ke sum ki tarah.

Horizontal axis par u aur vertical axis par v likha hai. Pehle step mein ek badhta hua curve point (u1, v1) se (u2, v2) tak jaata hai. Doosre step mein curve aur u-axis ke beech, u1 se u2 tak ka region shaded hai aur us par integral of v du likha hai. Teesre step mein curve aur v-axis ke beech, v1 se v2 tak ka region alag hatch se shaded hai aur us par integral of u dv likha hai. Chauthe step mein dono shaded regions saath dikhte hain: woh u2 v2 area wale bade rectangle ko bhar dete hain, sirf corner wala chhota rectangle u1 v1 bachta hai, isliye dono integrals ka sum hai u2 v2 minus u1 v1.

Example 17NCERT Example 17 · p. 260

First function chunna

Nikaalo $\int x \sin 2 x d x .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

$u$ aisa chuno jo differentiate karne par simple ho jaaye. Logarithms aur inverse trigonometric functions algebraic ban jaate hain, aur $x$ ki powers kam ho jaati hain. Exponentials aur sines aur cosines utne hi complicated rehte hain, isliye woh achhe second functions hain. Yaad rakhne ka ek common tareeka hai yeh order: inverse trig, log, algebraic, trig, exponential: jo type pehle aata hai, wahi aksar first function hota hai.
$\int g d x$ mein constant ki zaroorat nahi. Agar Example 17 mein hum $v = - \frac{1}{2} \cos 2 x + k$ lein, to extra terms hain $u v$ se $k x$ aur integral se $- \int k d x = - k x$ , aur yeh cancel ho jaate hain.
Parts hamesha madad nahi karta. Isse aisa second function chahiye jise hum integrate kar sakein, aur aisa naya integral jo simple ho. $\int \frac{\sin x}{x} d x$ ke liye koi choice kaam nahi karti: is integral ka elementary functions mein koi answer hai hi nahi.

Example 18NCERT Example 18 · p. 261

Second function ke roop mein 1

Nikaalo $\int \log (2 x + 1) d x,$ $x > - \frac{1}{2} .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 19NCERT Example 19 · p. 261

Power times exponential

Nikaalo $\int x e^{2 x} d x .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 20NCERT Example 20 · p. 261

Second function ko substitution chahiye

Nikaalo $\int x^{3} e^{x^{2}} d x .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 21NCERT Example 21 · p. 262

Jab integral wapas aa jaaye

Nikaalo $I = \int e^{- x} \cos 2 x d x .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

7.6.1 ∫eˣ[f(x) + f′(x)] dx type ka integral

TheoremProduct rule wala pattern

Agar $f$ differentiable hai, to $\int e^{x} [f (x) + f^{'} (x)] d x = e^{x} f (x) + C .$

Skill hai pattern pehchaanna: bracket ke andar ek term doosre ka derivative hona chahiye.

Example 22NCERT Example 22 · p. 263

f aur f′ pehchaanna

Nikaalo: (i) $\int e^{x} (x^{3} + 3 x^{2}) d x$ (ii) $\int e^{x} \frac{x^{2} + 3 x + 3}{(x + 2)^{2}} d x .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Practice · 5 questions

Practice 7.6: answers ke saath

1
Short answerEasyNCERT Exercise 7.6, Q1–Q8
By parts integrate karo.
1. $x \cos x$
  Solution dikhao
  Answer $x \sin x + \cos x + C$
  $u = x,$ $v = \sin x$ : $x \sin x - \int \sin x d x .$
2. $x \cos 4 x$
  Solution dikhao
  Answer $\frac{x}{4} \sin 4 x + \frac{1}{16} \cos 4 x + C$
  $u = x,$ $v = \frac{1}{4} \sin 4 x$ : $\frac{x}{4} \sin 4 x - \frac{1}{4} \int \sin 4 x d x .$
2
Short answerMediumNCERT Exercise 7.6, Q9–Q15
By parts integrate karo.
1. $x \cot^{- 1} x$
  Solution dikhao
  Answer $\frac{x^{2}}{2} \cot^{- 1} x + \frac{x}{2} - \frac{1}{2} \tan^{- 1} x + C$
  $u = \cot^{- 1} x$ ( $u^{'} = - \frac{1}{1 + x^{2}}$ ), $v = \frac{x^{2}}{2}$ : $\frac{x^{2}}{2} \cot^{- 1} x + \frac{1}{2} \int \frac{x^{2}}{1 + x^{2}} d x .$
  $\int \frac{x^{2}}{1 + x^{2}} d x = x - \tan^{- 1} x .$
2. $(\cos^{- 1} x)^{2}$
  Solution dikhao
  Answer $x (\cos^{- 1} x)^{2} - 2 \sqrt{1 - x^{2}} \cos^{- 1} x - 2 x + C$
  $u = (\cos^{- 1} x)^{2},$ $v = x$ : $x (\cos^{- 1} x)^{2} + 2 \int \frac{x \cos^{- 1} x}{\sqrt{1 - x^{2}}} d x .$
  Naye integral mein $\cos^{- 1} x$ ko first lo aur $\frac{x}{\sqrt{1 - x^{2}}}$ ko second (iska integral hai $- \sqrt{1 - x^{2}}$ ): yeh barabar hai $- \sqrt{1 - x^{2}} \cos^{- 1} x - x .$
3
Short answerMediumNCERT Exercise 7.6, Q16–Q22
Yeh formula use karo: $\int e^{x} [f (x) + f^{'} (x)] d x = e^{x} f (x) + C,$ ya do baar parts.
1. $e^{x} (\cos x - \sin x)$
  Solution dikhao
  Answer $e^{x} \cos x + C$
  $f (x) = \cos x,$ $f^{'} (x) = - \sin x .$
2. $\frac{(x + 2) e^{x}}{(x + 3)^{2}}$
  Solution dikhao
  Answer $\frac{e^{x}}{x + 3} + C$
  $\frac{x + 2}{(x + 3)^{2}} = \frac{1}{x + 3} - \frac{1}{(x + 3)^{2}};$ $f = \frac{1}{x + 3}$ lene par doosra term hai $f^{'} .$
Quick answerEasyNCERT Exercise 7.6, Q23
$\int x^{3} e^{x^{4}} d x$ barabar hai
$\frac{1}{4} e^{x^{4}} + C$ $\frac{1}{3} e^{x^{4}} + C$ $\frac{1}{4} e^{x^{3}} + C$ $\frac{x^{4}}{4} e^{x^{4}} + C$
Answer dekho
Answer (A) $\frac{1}{4} e^{x^{4}} + C$
$t = x^{4},$ lene par $d t = 4 x^{3} d x,$ isliye integral hai $\frac{1}{4} \int e^{t} d t = \frac{1}{4} e^{x^{4}} + C .$
Quick answerEasyNCERT Exercise 7.6, Q24
$\int e^{x} (\cot x - {c o s e c}^{2} x) d x$ barabar hai
$- e^{x} \cot x + C$ $e^{x} c o s e c x + C$ $e^{x} \tan x + C$ $e^{x} \cot x + C$
Answer dekho
Answer (D) $e^{x} \cot x + C$
$f = \cot x$ ka derivative hai $f^{'} = - {c o s e c}^{2} x,$ isliye bracket hai $f + f^{'}$ aur integral hai $e^{x} \cot x + C .$

7.6.2 Kuch aur types ke integrals

Teen square-root integrands hamari list poori karte hain. Yahan $a > 0.$

Table 7.5Quadratics ke square roots ke integralsNCERT 7.6.2 (i)–(iii) · p. 264

	Integral ( $C$ jodo)
(i)	$\int \sqrt{x^{2} - a^{2}} d x = \frac{x}{2} \sqrt{x^{2} - a^{2}} - \frac{a^{2}}{2} \log ∣ x + \sqrt{x^{2} - a^{2}} ∣$
(ii)	$\int \sqrt{x^{2} + a^{2}} d x = \frac{x}{2} \sqrt{x^{2} + a^{2}} + \frac{a^{2}}{2} \log ∣ x + \sqrt{x^{2} + a^{2}} ∣$
(iii)	$\int \sqrt{a^{2} - x^{2}} d x = \frac{x}{2} \sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2} \sin^{- 1} \frac{x}{a}$

Formula (iii) ke peeche ek picture hai. $y = \sqrt{a^{2} - x^{2}}$ ka graph radius $a,$ wale circle ka upper half hai, isliye $0 \leq x \leq a$ ke liye uske neeche $0$ se $x$ tak ka area ek right triangle aur ek circular sector mein bant jaata hai (Fig. 7.8).

3/3Sector ka angle $θ$ hai, jahan $\sin θ = \frac{x}{a}$ : area $\frac{a^{2}}{2} θ = \frac{a^{2}}{2} \sin^{- 1} \frac{x}{a} .$

Fig. 7.8Circle ke neeche ka area: ek triangle plus ek sector.

First quadrant mein radius a ka quarter circle, jiska centre origin O par hai. Pehle step mein x barabar 0 aur x par ek vertical line ke beech arc ke neeche ka region shaded hai, aur vertical line ka top circle par point P hai. Doosre step mein O, vertical line ka foot, aur P corners wala triangle highlight hota hai; iska base x hai aur height a squared minus x squared ka square root hai. Teesre step mein bacha hua tukda, vertical axis aur radius OP ke beech ka sector jiska angle O par theta hai, highlight hota hai; sin theta barabar x over a.

Dono areas jodne par bilkul (iii) ki right-hand side milti hai, aur yeh $x = 0$ par zero ho jaati hai, jaise $0$ se $0$ tak ka area hona hi chahiye. Ek saaf proof, jo valid hai jab bhi $∣ x ∣ \leq a,$ yeh hai ki substitute karo $x = a \sin θ$ : tab $\int \sqrt{a^{2} - x^{2}} d x = a^{2} \int \cos^{2} θ d θ = \frac{a^{2}}{2} (θ + \sin θ \cos θ),$ jo wahi expression hai. (ii) ke liye hum $1$ ko second function lekar parts use karte hain:

DerivationFormula (ii) ka proof

Maan lo $I = \int \sqrt{x^{2} + a^{2}} d x .$ $\sqrt{x^{2} + a^{2}}$ ko first aur $1$ ko second lekar by parts integrate karne par,

\begin{aligned} I & = x \sqrt{x^{2} + a^{2}} \\ - \int \frac{x^{2} d x}{\sqrt{x^{2} + a^{2}}} \end{aligned}

Aakhri numerator mein likho $x^{2} = (x^{2} + a^{2}) - a^{2}$ . Integral do hisson mein bant jaata hai: khud $I$ , aur $a^{2}$ times formula (6), jise hum short mein likhte hain $L = \log ∣ x + \sqrt{x^{2} + a^{2}} ∣$ :

\begin{aligned} I & = x \sqrt{x^{2} + a^{2}} \\ - I + a^{2} L \end{aligned}

To $2 I = x \sqrt{x^{2} + a^{2}} + a^{2} L,$ jisse (ii) milta hai. Formula (i) bilkul isi tarah formula (4) use karke aata hai, aur substitutions $x = a \tan θ,$ $x = a \sec θ$ se bhi yahi results milte hain.

Example 23NCERT Example 23 · p. 265

Square complete karo, phir (ii) use karo

Nikaalo $\int \sqrt{x^{2} + 4 x + 8} d x .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 24NCERT Example 24 · p. 265

Square complete karo, phir (iii) use karo

Nikaalo $\int \sqrt{5 + 4 x - x^{2}} d x .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

2 mein se 1

Square complete karne ke baad kaun sa formula

\int \sqrt{8 - 2 x - x^{2}} d x

ko poora karta hai?

(i), jahan

a = 3

(ii), jahan

a = 3

(iii), jahan

a = 3

(iii), jahan

a = 9

Answer dekho

Answer (C) (iii), jahan $a = 3$

8 - 2 x - x^{2} = 9 - (x + 1)^{2}

: ek constant minus ek square, isliye (iii), jahan

a = 3

(na ki

a = 9

a^{2} = 9

Practice · 4 questions

Practice 7.7: answers ke saath

1
Short answerMediumNCERT Exercise 7.7, Q1–Q8
Formulas (i)–(iii) se har function integrate karo.
1. $\sqrt{9 - x^{2}}$
  Solution dikhao
  Answer $\frac{x}{2} \sqrt{9 - x^{2}} + \frac{9}{2} \sin^{- 1} \frac{x}{3} + C$
  Formula (iii), jahan $a = 3.$
2. $\sqrt{4 - 9 x^{2}}$
  Solution dikhao
  Answer $\frac{x}{2} \sqrt{4 - 9 x^{2}} + \frac{2}{3} \sin^{- 1} \frac{3 x}{2} + C$
  $t = 3 x,$ $d x = \frac{d t}{3}$ : $\frac{1}{3} [\frac{t}{2} \sqrt{4 - t^{2}} + 2 \sin^{- 1} \frac{t}{2}] .$
Short answerMediumNCERT Exercise 7.7, Q9
Integrate karo $\sqrt{4 + \frac{x^{2}}{4}} .$
Solution dikhao
Answer $\frac{x}{4} \sqrt{x^{2} + 16} + 4 \log ∣ x + \sqrt{x^{2} + 16} ∣ + C$
$\sqrt{4 + \frac{x^{2}}{4}} = \frac{1}{2} \sqrt{x^{2} + 16} .$
Formula (ii), jahan $a = 4$ : $\frac{1}{2} [\frac{x}{2} \sqrt{x^{2} + 16} + 8 \log ∣ x + \sqrt{x^{2} + 16} ∣] + C .$
Quick answerEasyNCERT Exercise 7.7, Q10
$\int \sqrt{x^{2} + 9} d x$ barabar hai
$\frac{2}{3} (x^{2} + 9)^{3 / 2} + C$ $\frac{x}{2} \sqrt{x^{2} + 9} + \frac{9}{2} \log ∣ x + \sqrt{x^{2} + 9} ∣ + C$ $\frac{x}{2} \sqrt{x^{2} + 9} + \frac{9}{2} \sin^{- 1} \frac{x}{3} + C$ $\frac{x}{2} \sqrt{x^{2} + 9} + \frac{3}{2} \log ∣ x + \sqrt{x^{2} + 9} ∣ + C$
Answer dekho
Answer (B) $\frac{x}{2} \sqrt{x^{2} + 9} + \frac{9}{2} \log ∣ x + \sqrt{x^{2} + 9} ∣ + C$
Formula (ii): $\frac{x}{2} \sqrt{x^{2} + a^{2}} + \frac{a^{2}}{2} \log ∣ x + \sqrt{x^{2} + a^{2}} ∣$ , jahan $a^{2} = 9.$ Option A root ko power ki tarah treat karta hai, jaise power of $x;$ option D use karta hai $a$ , jabki chahiye $a^{2} .$
Quick answerMediumNCERT Exercise 7.7, Q11
$\int \sqrt{x^{2} + 6 x + 5} d x$ barabar hai
$\frac{x + 3}{2} \sqrt{x^{2} + 6 x + 5} + 2 \log ∣ x + 3 + \sqrt{x^{2} + 6 x + 5} ∣ + C$ $\frac{x + 3}{2} \sqrt{x^{2} + 6 x + 5} - 4 \log ∣ x + 3 + \sqrt{x^{2} + 6 x + 5} ∣ + C$ $\frac{x + 3}{2} \sqrt{x^{2} + 6 x + 5} - 2 \log ∣ x + 3 + \sqrt{x^{2} + 6 x + 5} ∣ + C$ $\frac{x - 3}{2} \sqrt{x^{2} + 6 x + 5} - 2 \log ∣ x - 3 + \sqrt{x^{2} + 6 x + 5} ∣ + C$
Answer dekho
Answer (C) $\frac{x + 3}{2} \sqrt{x^{2} + 6 x + 5} - 2 \log ∣ x + 3 + \sqrt{x^{2} + 6 x + 5} ∣ + C$
Formula (i), jahan $y = x + 3,$ $a = 2$ : log term mein aata hai $- \frac{a^{2}}{2} = - 2.$ Option A mein formula (ii) wala sign hai; option B aadha karna bhool jaata hai is term ko: $a^{2} .$

7.7 Definite Integral

Ab hum Section 7.1 ke doosre sawaal par aate hain: graph ke neeche ka area. Maan lo $f$ poore interval par $x$ -axis ke upar rehta hai, aur interval hai $[a, b] .$ Jo region $y = f (x),$ axis, aur vertical lines $x = a$ aur $x = b$ ke beech hai, uska kuch area hai; woh number, functions ki family nahi, $a$ se $b$ tak $f$ ka definite integral hai, jise likhte hain $\int_{a}^{b} f (x) d x .$ Numbers $a$ aur $b$ limits of integration hain: $a$ integral sign ke neeche, $b$ upar.

Definition

Definite integral

Maan lo $f$ continuous hai aur interval hai $[a, b],$ to $\int_{a}^{b} f (x) d x$ woh signed area hai jo $f$ ke graph aur $x$ -axis ke beech, $x = a$ se $x = b$ tak hai: axis ke upar ka area positive gina jaata hai, neeche ka negative. Agar $F$ , $f$ ka antiderivative hai aur interval hai $[a, b],$ to iski value hai $F (b) - F (a) .$

Example $\int_{0}^{2} x d x$ us triangle ka area hai jiska base hai $2$ aur height hai $2,$ yaani area hai $2;$ aur $\frac{2^{2}}{2} - \frac{0^{2}}{2} = 2$ bhi.

Aisa area kaise nikaalein? Ek tareeka hai approximate karna: $[a, b]$ ko $n$ patli strips mein kaato, har strip ki jagah ek rectangle rakho, aur sab jodo. Jaise-jaise $n$ badhta hai, rectangles curve se chipakte jaate hain aur total ek value par tik jaata hai, jise limit of a sum kehte hain (Fig. 7.9). Doosra tareeka sums ko chhod deta hai: ek antiderivative $F$ dhoondho aur nikaalo $F (b) - F (a) .$ Section 7.8 samjhata hai ki dono hamesha match kyun karte hain, aur uske baad hum doosra, kahin zyada tez, raasta use karte hain.

2/3Aath rectangles: gaps chhote ho jaate hain.

Fig. 7.9Patle se patle rectangles jodne par hum curve ke neeche ke area ke paas pahunchte hain.

x-axis par a se b tak ke interval ke upar ek badhta hua curve, jiske neeche ka asli area outline kiya gaya hai. Pehle step mein curve ke neeche barabar width ke chaar rectangles hain, har ek apne left edge par curve jitna ooncha, aur curve ke neeche gaps dikhte hain. Doosre step mein aath patle rectangles hain aur gaps chhote hain. Teesre step mein battees bahut patle rectangles hain aur woh region ko lagbhag poora bhar dete hain.

InteractiveArea explorer

Ek interactive graph jismein chaar functions mein se chun sakte ho: x squared, x ka square root, sine x aur one over x. Sliders interval ke ends a aur b set karte hain. Sums mode mein ek slider strips ki ginti 1 se 64 tak set karta hai aur buttons left, right ya midpoint rectangles chunte hain; panel rectangle total, antiderivative se nikli exact value F of b minus F of a, aur dono ka difference dikhata hai, jo strips patli hone par kam hota jaata hai. Area function mode mein ek slider point x ko a se b tak le jaata hai; curve ke neeche a se x tak ka region shaded hota hai, neeche ek doosra chhota graph area A ko x ke against plot karta hai jaise-jaise woh banta hai, aur us graph ke current point par ek chhote tangent ka slope f of x ke barabar hota hai, aur dono numbers saath-saath dikhte hain.

Fig. 7.10Ek function aur ek interval chuno. Sums mode mein strips ki ginti badlo aur rectangle total ko compare karo is se:

F (b) - F (a);

Area function mode mein

x

ko move karo aur dekho ki

A (x)

kis rate se badhta hai:

f (x) .

Quick check

3 mein se 1

\int_{- 1}^{0} x d x

hai

\frac{1}{2},

ek triangle ka area

0,

kyunki interval

0

par khatam hota hai

- \frac{1}{2},

kyunki region

x

-axis ke neeche haiundefined, kyunki wahan

x

negative hai

Answer dekho

Answer (C) $- \frac{1}{2},$ kyunki region $x$ -axis ke neeche hai

[- 1, 0]

par line

y = x

axis ke neeche hai, isliye

\frac{1}{2}

area wala triangle negative gina jaata hai:

[\frac{x^{2}}{2}]_{- 1}^{0} = 0 - \frac{1}{2} .

7.8 Fundamental Theorem of Calculus

7.8.1 Area function

Left end $a$ ko fixed rakho aur right end ko axis ke saath khiskao. Right end ki har position $x$ ek alag area gherti hai, us curve ke neeche jo hai $y = f (t),$ isliye area ek aisi quantity ban jaata hai jo depend karti hai is par: $x .$

Definition

Area function

Maan lo $f$ continuous hai aur interval hai $[a, b],$ to area function hai $A (x) = \int_{a}^{x} f (t) d t$ , jahan $a \leq x \leq b .$ Yeh graph ke neeche $a$ se $x$ tak ka area hai (signed, agar $f$ ka sign badle), aur $A (a) = 0.$

Example $f (t) = 2$ aur $a = 0$ ke liye region ek rectangle hai, isliye $A (x) = 2 x .$

Integral ke andar ke variable ko humne $t$ likha taaki woh moving end ke saath confuse na ho, jo hai $x;$ andar use kiya gaya letter value nahi badalta (yeh Section 7.10 mein property P0 hai). Fig. 7.10 ka Area function mode try karo: jaise-jaise $x$ right ki taraf jaata hai, $A (x)$ wahan tezi se badhta hai jahan curve ooncha hai aur wahan dheere jahan neecha hai.

7.8.2 Integral calculus ka first fundamental theorem

Theorem 1First fundamental theorem

Maan lo $f$ continuous hai $[a, b]$ par aur $A (x) = \int_{a}^{x} f (t) d t .$ Tab $A^{'} (x) = f (x)$ , har us $x$ ke liye jo is interval mein hai: $[a, b] .$

Poora proof is book ke level se aage hai, lekin wajah Fig. 7.11 mein dikhti hai. Jab $x$ thoda sa badhta hai, maan lo $h,$ to area ek patli strip jitna badhta hai jiski width $h$ hai aur height hai lagbhag $f (x),$ kyunki $f$ continuous hai. Isliye $A (x + h) - A (x) \approx f (x) h,$ aur

\begin{array}{c} \frac{A (x + h) - A (x)}{h} \\ \approx f (x) \end{array}

Error isliye aata hai kyunki strip ka top bilkul flat nahi hota, aur yeh $0$ tak chhota ho jaata hai jab ho $h \to 0.$ Left side ki limit hai $A^{'} (x) .$

3/3Strip lagbhag ek rectangle hai jiski height $f (x)$ aur width hai $h,$ isliye $\frac{A (x + h) - A (x)}{h} \to f (x) .$

Fig. 7.11Area function ka derivative f(x) kyun hai.

x-axis ke upar ek curve. Pehle step mein a se x tak uske neeche ka region shaded hai aur us par A of x likha hai. Doosre step mein x ek chhoti si doori h se right ki taraf jaata hai aur x aur x plus h ke beech ki nayi patli strip alag pattern mein shaded hoti hai. Teesre step mein strip ke upar width h aur height f of x ka ek rectangle outline hota hai, jo strip se lagbhag bilkul match karta hai; sirf top par ek chhota sa gap hi farq hai.

Words mein: area jis rate se badhta hai, woh curve ki height hai. Isliye area function ek antiderivative hai, jiska derivative hai $f .$ Yahi Section 7.1 ki dono problems ke beech ka pul hai.

7.8.3 Integral calculus ka second fundamental theorem

Theorem 2Second fundamental theorem

Maan lo $f$ continuous hai $[a, b]$ par aur $F$ , $f$ ka koi bhi antiderivative hai, us interval par jo hai $[a, b] .$ Tab $\int_{a}^{b} f (x) d x = F (b) - F (a) .$

Definite integral evaluate karna

\begin{aligned} \int_{a}^{b} f (x) d x \\ = [F (x)]_{a}^{b} \\ = F (b) - F (a) \end{aligned}

Koi bhi antiderivative chalega. $F$ ki jagah $F + C$ lene par milta hai $[F (b) + C] - [F (a) + C],$ aur $C$ cancel ho jaata hai. Isliye definite integral mein hum kabhi $C$ nahi likhte.
Sums ki zaroorat nahi. Value antiderivative se aati hai, rectangles jodne se nahi. Mushkil hissa pehle jaisa hi hai: dhoondhna $F .$
Integrand poore $[a, b]$ par continuous hona chahiye. Ise ignore karne se bekaar result aata hai. Maan lo integral hai $\int_{- 1}^{2} \frac{d x}{x^{2}},$ to bina soche $F (x) = - \frac{1}{x}$ use karne par milta hai $- \frac{1}{2} - 1 = - \frac{3}{2},$ yaani positive integrand ka negative “area”. Theorem yahan laagu nahi hota, kyunki $\frac{1}{x^{2}}$ defined nahi hai yahan: $x = 0,$ jo is interval ke andar hai: $[- 1, 2] .$

Step by step

Definite integral ka working rule

Step 1 / 3

Antiderivative

Is chapter ke kisi bhi method se nikaalo $\int f (x) d x = F (x),$ bina constant ke.

1 / 3

Example 25NCERT Example 25 · p. 269

Chaar definite integrals

Evaluate karo: (i) $\int_{1}^{4} \sqrt{x} d x$ (ii) $\int_{0}^{1} \frac{x^{2}}{(1 + x^{3})^{2}} d x$ (iii) $\int_{0}^{1} \frac{d x}{(x + 2) (x + 3)}$ (iv) $\int_{0}^{π / 6} \cos^{2} 3 t \sin 3 t d t .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Practice · 5 questions

Practice 7.8: answers ke saath

1
Short answerEasyNCERT Exercise 7.8, Q1–Q8
Har definite integral evaluate karo.
1. $\int_{- 2}^{1} (2 x + 3) d x$
  Solution dikhao
  Answer $6$
  $[x^{2} + 3 x]_{- 2}^{1} = (1 + 3) - (4 - 6) = 6.$
2. $\int_{1}^{e} \frac{2}{x} d x$
  Solution dikhao
  Answer $2$
  $[2 \log x]_{1}^{e} = 2 - 0.$
2
Short answerMediumNCERT Exercise 7.8, Q9–Q15
Har definite integral evaluate karo.
1. $\int_{0}^{1 / 2} \frac{d x}{\sqrt{1 - x^{2}}}$
  Solution dikhao
  Answer $\frac{π}{6}$
  $[\sin^{- 1} x]_{0}^{1 / 2} .$
2. $\int_{0}^{2} \frac{d x}{4 + x^{2}}$
  Solution dikhao
  Answer $\frac{π}{8}$
  $[\frac{1}{2} \tan^{- 1} \frac{x}{2}]_{0}^{2} = \frac{1}{2} \cdot \frac{π}{4} .$
3
Short answerHardNCERT Exercise 7.8, Q16–Q20
Har definite integral evaluate karo.
1. $\int_{1}^{2} \frac{x^{2}}{x^{2} + 3 x + 2} d x$
  Solution dikhao
  Answer $1 + \log \frac{3}{2} - 4 \log \frac{4}{3}$
  Divide karo aur todo: $\frac{x^{2}}{(x + 1) (x + 2)} = 1 + \frac{1}{x + 1} - \frac{4}{x + 2} .$
  Antiderivative hai $x + \log (x + 1) - 4 \log (x + 2);$ $1$ se $2$ tak: $1 + \log \frac{3}{2} - 4 \log \frac{4}{3} .$
2. $\int_{0}^{π / 3} (\sec^{2} x + 2 x + 1) d x$
  Solution dikhao
  Answer $\sqrt{3} + \frac{π^{2}}{9} + \frac{π}{3}$
  $[\tan x + x^{2} + x]_{0}^{π / 3} = \sqrt{3} + \frac{π^{2}}{9} + \frac{π}{3} .$
Quick answerEasyNCERT Exercise 7.8, Q21
$\int_{1 / \sqrt{3}}^{1} \frac{d x}{1 + x^{2}}$ barabar hai
$\frac{π}{3}$ $\frac{π}{6}$ $\frac{π}{4}$ $\frac{π}{12}$
Answer dekho
Answer (D) $\frac{π}{12}$
$\tan^{- 1} 1 - \tan^{- 1} \frac{1}{\sqrt{3}} = \frac{π}{4} - \frac{π}{6} = \frac{π}{12} .$
Quick answerMediumNCERT Exercise 7.8, Q22
$\int_{0}^{1 / 2} \frac{d x}{1 + 4 x^{2}}$ barabar hai
$\frac{π}{8}$ $\frac{π}{4}$ $\frac{π}{2}$ $\frac{π}{16}$
Answer dekho
Answer (A) $\frac{π}{8}$
$\int \frac{d x}{1 + 4 x^{2}} = \frac{1}{2} \tan^{- 1} 2 x;$ $0$ se $\frac{1}{2}$ tak yeh hai $\frac{1}{2} \cdot \frac{π}{4} = \frac{π}{8} .$ Option B yeh factor bhool jaata hai: $\frac{1}{2} .$

7.9 Substitution se Definite Integrals ka Evaluation

Substitution ke saath $\int_{a}^{b} f (x) d x$ evaluate karne ke liye hum indefinite integral nikaal kar wapas la sakte hain variable $x,$ aur phir limits use kar sakte hain $a$ aur $b .$ Aksar zyada tez yeh hota hai ki variable ke saath limits bhi badal do aur kabhi wapas na lautao variable $x,$ jaisa Fig. 7.12 dikhata hai.

2/3 $t = x^{3} + 1$ lene par $x = 0$ ban jaata hai $t = 1$ aur $x = 2$ ban jaata hai $t = 9.$

Fig. 7.12Jab variable badalta hai, limits bhi uske saath badalti hain.

Do horizontal number lines, upar x aur neeche t. Pehle step mein x line par 0 se 2 tak ka interval highlight hota hai. Doosre step mein t barabar x cubed plus 1 likhe arrows x line ke points 0, 1 aur 2 ko t line ke points 1, 2 aur 9 par le jaate hain; arrows phail jaate hain kyunki map stretch karta hai. Teesre step mein t line par 1 se 9 tak ka interval highlight hota hai aur integral ko limits 1 aur 9 ke saath dobara likha jaata hai.

Step by step

Definite integral mein substitution ka working rule

Step 1 / 3

Substitute karo

$t = g (x)$ chuno aur likho $d t = g^{'} (x) d x,$ bilkul indefinite integral ki tarah.

1 / 3

Example 26NCERT Example 26 · p. 272

Do raaste, ek value

Evaluate karo $\int_{0}^{2} 3 x^{2} \sqrt{x^{3} + 1} d x .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 27NCERT Example 27 · p. 272

Naye variable ke roop mein inverse sine

Evaluate karo $\int_{0}^{1 / 2} \frac{\sin^{- 1} x}{\sqrt{1 - x^{2}}} d x .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

\int_{1}^{2} x (x^{2} - 1)^{3} d x

mein rakho

t = x^{2} - 1.

Nayi limits hain

1

2

tak

1

4

tak

0

1

tak

0

3

tak

Answer dekho

Answer (D) $0$ se $3$ tak

x = 1

se milta hai

t = 0

aur

x = 2

se milta hai

t = 3.

Purani limits

1

aur

2

hi rakh lena (option A) sabse common galti hai.

Practice · 3 questions

Practice 7.9: answers ke saath

1
Short answerMediumNCERT Exercise 7.9, Q1–Q8
Limits badalte hue substitution se evaluate karo.
1. $\int_{0}^{1} \frac{x^{2}}{x^{3} + 1} d x$
  Solution dikhao
  Answer $\frac{1}{3} \log 2$
  $t = x^{3} + 1$ : limits $1$ se $2;$ $\frac{1}{3} \int_{1}^{2} \frac{d t}{t} .$
2. $\int_{0}^{π / 2} \sqrt{\cos ϕ} \sin^{3} ϕ d ϕ$
  Solution dikhao
  Answer $\frac{8}{21}$
  $\sin^{3} ϕ = (1 - \cos^{2} ϕ) \sin ϕ;$ $t = \cos ϕ,$ limits $1$ se $0$ : $\int_{0}^{1} t^{1 / 2} (1 - t^{2}) d t = \frac{2}{3} - \frac{2}{7} .$
Quick answerHardNCERT Exercise 7.9, Q9
$\int_{1}^{2} \frac{\sqrt{x^{2} - 1}}{x^{4}} d x$ ki value hai
$\frac{\sqrt{3}}{4}$ $\frac{3}{8}$ $0$ $\frac{\sqrt{3}}{8}$
Answer dekho
Answer (D) $\frac{\sqrt{3}}{8}$
$d t = \frac{2}{x^{3}} d x;$ limits $0$ se $\frac{3}{4} .$ To integral hai $\frac{1}{2} \int_{0}^{3 / 4} t^{1 / 2} d t = \frac{1}{3} (\frac{3}{4})^{3 / 2} = \frac{1}{3} \cdot \frac{3 \sqrt{3}}{8} = \frac{\sqrt{3}}{8} .$
Quick answerEasyNCERT Exercise 7.9, Q10
Agar $f (x) = \int_{0}^{x} t^{2} \cos t d t,$ to $f^{'} (x)$ hai
$2 x \cos x - x^{2} \sin x$ $x^{2} \cos x$ $x^{2} \sin x$ $x^{2} \cos x - 1$
Answer dekho
Answer (B) $x^{2} \cos x$
Theorem 1 se $\int_{0}^{x} g (t) d t$ ka derivative hai $g (x),$ yahan $x^{2} \cos x .$ Option A iski jagah integrand ko differentiate kar deta hai.

7.10 Definite Integrals ki Kuch Properties

Kuch definite integrals antiderivative ke through karna mushkil ya namumkin hota hai, lekin interval ya integrand ki symmetry use karte hi woh aasaan ho jaate hain. Table 7.6 ki aath properties yahi tools hain. Har ek second fundamental theorem se ya kisi substitution se nikalti hai.

Table 7.6Definite integrals ki properties (yahan

\int_{a}^{b} f

short form hai is integral ka:

\int_{a}^{b} f (x) d x

)NCERT P0–P7 · p. 273

	Property
P0	$\int_{a}^{b} f (x) d x = \int_{a}^{b} f (t) d t$
P1	$\int_{a}^{b} f = - \int_{b}^{a} f,$ and $\int_{a}^{a} f = 0$
P2	$\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$
P3	$\int_{a}^{b} f (x) d x$ nahi badalta agar $x$ ki jagah $a + b - x$ rakh do
P4	$[0, a]$ ke liye P3: $x$ ki jagah $a - x$ rakho
P5	$\int_{0}^{2 a} f = \int_{0}^{a} [f (x) + f (2 a - x)] d x$
P6	Jab $f (2 a - x) = f (x)$ ( $x = a$ ke around mirror symmetry), to $\int_{0}^{2 a} f = 2 \int_{0}^{a} f .$ Jab $f (2 a - x) = - f (x),$ to dono halves cancel hokar $0$ ho jaate hain
P7	$\int_{- a}^{a} f$ hota hai $2 \int_{0}^{a} f$ jab even ho $f,$ aur $0$ jab odd ho $f$

PropertyP0, P1, P2: naam badalna, ulat dena, todna

Variable ka naam value par asar nahi daalta (P0). Limits ulatne se sign ulat jaata hai, aur zero width wala interval deta hai $0$ (P1). Integral ko kisi bhi point $c$ par toda ja sakta hai (P2).

PropertyP3 aur P4: interval ko reflect karna

$\int_{a}^{b}$ ke andar $x$ ki jagah $a + b - x$ rakhne se value nahi badalti (P3); interval $[0, a]$ ke liye replacement hai $x \to a - x$ (P4).

$\int_{a}^{b} f (x) d x$ mein rakho $x = a + b - t,$ to $d x = - d t .$ Jab $x = a,$ to $t = b;$ jab $x = b,$ to $t = a .$ Integral ban jaata hai $- \int_{b}^{a} f (a + b - t) d t,$ jo barabar hai $\int_{a}^{b} f (a + b - t) d t$

P1 se limits swap karke; P0 se hum phir $x$ likh sakte hain, is variable ki jagah: $t .$ $a = 0$ rakhne par (aur upper limit ko $a$ kehne par) P4 milta hai. Geometry mein, $f (a + b - x)$ interval ke beech se guzarti line mein $f (x)$ ki mirror image hai, isliye woh utna hi area gherta hai (Fig. 7.13).

PropertyP5 aur P6: interval ko fold karna

$[0, 2 a]$ par integral barabar hai $\int_{0}^{a} [f (x) + f (2 a - x)] d x$ (P5). Isliye yeh $\int_{0}^{a} f$ ka double hota hai jab $f$ symmetric ho is line ke around: $x = a,$ aur zero hota hai jab $f$ antisymmetric ho is line ke around: $x = a$ (P6).

PropertyP7: even aur odd functions

Interval $[- a, a]$ symmetric hai is point ke around: $0,$ aur aise interval par even integrand $[0, a]$ wale integral ka double deta hai aur odd integrand deta hai $0.$

2/3Odd function ke areas barabar aur ulte hote hain: $\int_{- a}^{a} f = 0.$

Fig. 7.13Symmetry se bina antiderivative ke bhi integral evaluate ho sakta hai.

Pehle step mein ek even curve, jo y-axis ke around symmetric hai, minus a aur a ke beech axis ke left aur right par barabar shaded areas rakhta hai, isliye total ek side ka double hai. Doosre step mein ek odd curve origin se guzarta hai; right ka shaded area x-axis ke upar hai aur left ka shaded area usi shape ka uske neeche hai, plus aur minus marked, isliye woh cancel ho jaate hain. Teesre step mein 0 se a tak ke interval par ek curve, a over 2 par vertical line mein uski mirror image ke saath, dashed line se bana hai; dono same area gherte hain, yahi property P4 hai.

Example 28NCERT Example 28 · p. 276

Modulus: jahan sign badle wahan todo

Evaluate karo $\int_{0}^{3} ∣ x^{2} - 3 x + 2 ∣ d x .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

3/3 $1$ aur $2$ par todo: $\frac{5}{6} + \frac{1}{6} + \frac{5}{6} = \frac{11}{6} .$

Fig. 7.14Modulus lene se negative tukda upar palat jaata hai.NCERT p. 276

x ki value 0 se 3 tak y barabar x squared minus 3x plus 2 ka graph. Pehle step mein curve height 2 se shuru hota hai, 1 par x-axis ko cross karta hai, 1.5 par minus one quarter tak neeche jaata hai, 2 par phir cross karta hai aur x barabar 3 par 2 tak upar jaata hai; 1 aur 2 ke beech ka hissa axis ke neeche shaded hai aur minus marked hai. Doosre step mein modulus liya jaata hai: dip axis ke upar ek chhote bump ki tarah reflect hota hai, dashed line se bana. Teesre step mein teeno tukde shaded hain aur un par unke areas likhe hain, five sixths, one sixth aur five sixths, jinka sum eleven sixths hai.

Example 29NCERT Example 29 · p. 276

Even integrand

Evaluate karo $\int_{- π / 3}^{π / 3} \cos^{2} x d x .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 30NCERT Example 30 · p. 277

Aage wale x se chhutkaara

Evaluate karo $I = \int_{0}^{π} x \sin x \cos^{2} x d x .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 31NCERT Example 31 · p. 277

Odd integrand

Evaluate karo $\int_{- 2}^{2} x^{3} e^{x^{2}} d x .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 32NCERT Example 32 · p. 278

Mirror image jodo

Evaluate karo $I = \int_{0}^{π / 2} \frac{d x}{1 + \tan^{3} x} .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 33NCERT Example 33 · p. 278

0 se shuru na hone wale interval par P3

Evaluate karo $I = \int_{2}^{5} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{7 - x}} d x .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 34NCERT Example 34 · p. 279

Ek mashhoor logarithmic integral

Evaluate karo $J = \int_{0}^{π} \log \sin x d x,$ aur isse nikaalo $\int_{0}^{π / 2} \log \sin x d x .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Practice · 6 questions

Practice 7.10: answers ke saath

1
Short answerMediumNCERT Exercise 7.10, Q1–Q7
Symmetry aur properties P0–P7 use karo; lambe antiderivative ki zaroorat nahi.
1. $\int_{0}^{π / 2} \sin^{2} x d x$
  Solution dikhao
  Answer $\frac{π}{4}$
  P4 se yeh barabar hai $\int_{0}^{π / 2} \cos^{2} x d x .$ Jodne par, $2 I = \int_{0}^{π / 2} (\sin^{2} x + \cos^{2} x) d x = \frac{π}{2} .$
2. $\int_{0}^{π / 2} \frac{\sqrt[3]{\sin x}}{\sqrt[3]{\sin x} + \sqrt[3]{\cos x}} d x$
  Solution dikhao
  Answer $\frac{π}{4}$
  P4 swap kar deta hai $\sin$ aur $\cos;$ dono forms jodne par milta hai $2 I = \frac{π}{2} .$
2
Short answerHardNCERT Exercise 7.10, Q8–Q14
Har integral evaluate karo.
1. $\int_{0}^{1} \frac{\log (1 + x)}{1 + x^{2}} d x$
  Solution dikhao
  Answer $\frac{π}{8} \log 2$
  Rakho $x = \tan θ$ : $I = \int_{0}^{π / 4} \log (1 + \tan θ) d θ .$
  P4 se, $1 + \tan (\frac{π}{4} - θ) = 1 + \frac{1 - \tan θ}{1 + \tan θ} = \frac{2}{1 + \tan θ},$ isliye $I = \int_{0}^{π / 4} [\log 2 - \log (1 + \tan θ)] d θ = \frac{π}{4} \log 2 - I .$
2. $\int_{0}^{1} x^{2} \sqrt{1 - x} d x$
  Solution dikhao
  Answer $\frac{16}{105}$
  P4: $\int_{0}^{1} (1 - x)^{2} \sqrt{x} d x = \int_{0}^{1} (x^{1 / 2} - 2 x^{3 / 2} + x^{5 / 2}) d x = \frac{2}{3} - \frac{4}{5} + \frac{2}{7} .$
3
Short answerMediumNCERT Exercise 7.10, Q15–Q18
Evaluate ya prove karo.
1. $\int_{0}^{π / 2} \frac{\sin^{2} x - \cos^{2} x}{2 + \sin x \cos x} d x$
  Solution dikhao
  Answer $0$
  P4 $\sin x$ aur $\cos x$ ko swap karta hai: numerator ka sign badal jaata hai, denominator ka nahi. Isliye $I = - I .$
2. $\int_{0}^{π} \log (1 - \cos x) d x$
  Solution dikhao
  Answer $- π \log 2$
  $1 - \cos x = 2 \sin^{2} \frac{x}{2},$ isliye $I = π \log 2 + 2 \int_{0}^{π} \log \sin \frac{x}{2} d x .$
  $x = 2 t$ lene par aakhri integral hai $2 \int_{0}^{π / 2} \log \sin t d t = - π \log 2$ (Example 34). To $I = π \log 2 - 2 π \log 2.$
ProofHardNCERT Exercise 7.10, Q19
Agar $f (a - x) = f (x)$ aur $g (x) + g (a - x) = 6$ , har us $x$ ke liye jo is interval mein hai: $[0, a],$ to dikhao ki $\int_{0}^{a} f (x) g (x) d x = 3 \int_{0}^{a} f (x) d x .$
Solution dikhao
Answer $I = 3 \int_{0}^{a} f (x) d x$
P4 se, $I = \int_{0}^{a} f (a - x) g (a - x) d x = \int_{0}^{a} f (x) [6 - g (x)] d x = 6 \int_{0}^{a} f (x) d x - I .$
To $2 I = 6 \int_{0}^{a} f (x) d x,$ jo result hai.
Quick answerEasyNCERT Exercise 7.10, Q20
$\int_{- 1}^{1} (x^{7} - x \cos x + \sin^{3} x + 2) d x$ ki value hai
$0$ $2$ $4$ $8$
Answer dekho
Answer (C) $4$
$x^{7},$ $x \cos x$ aur $\sin^{3} x$ odd hain, isliye inka contribution hai $0.$ Constant $2$ deta hai $\int_{- 1}^{1} 2 d x = 4.$
Quick answerMediumNCERT Exercise 7.10, Q21
$\int_{0}^{1} \log \frac{1 + 2 x}{3 - 2 x} d x$ ki value hai
$\log 3$ $0$ $1$ $2 \log 3$
Answer dekho
Answer (B) $0$
$x \to 1 - x,$ karne par $\frac{1 + 2 x}{3 - 2 x}$ ban jaata hai $\frac{3 - 2 x}{1 + 2 x},$ isliye integrand ka sign badal jaata hai: $I = - I,$ isliye $I = 0.$

Miscellaneous Examples

Asli integrals apne method ka label lagakar nahi aate. Neeche ke examples chapter ki techniques ko mix karte hain. Calculate karne se pehle Fig. 7.15 ke sawaalon se guzro: zyadatar integrands ek minute mein tay ho jaate hain.

4/4Kya yeh do alag tarah ke functions ka product hai? Integration by parts try karo.

Fig. 7.15Kisi bhi integrand se poochhne ke chaar sawaal, isi order mein.

Chaar panels ek ke neeche ek, neeche ki taraf arrows se jude. Pehle step mein sabse upar wala panel, standard form, table ke ek sample integrand ke saath highlight hota hai. Doosre step mein doosra panel, inner function aur uska derivative, sample integrand 2x cos of x squared aur word substitution ke saath highlight hota hai. Teesre step mein teesra panel, ratio of polynomials, ek sample rational function aur words partial fractions ke saath highlight hota hai. Chauthe step mein sabse neeche wala panel, do tarah ka product, x times e to the x aur words by parts ke saath highlight hota hai.

Example 35NCERT Example 35 · p. 281

Root ke andar substitution

Nikaalo $\int \sin 3 x \sqrt{5 - \cos 3 x} d x .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 36NCERT Example 36 · p. 281

Sabse badi power bahar nikaalo

Nikaalo $\int \frac{(x^{3} + x)^{1 / 3}}{x^{4}} d x,$ $x > 0.$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 37NCERT Example 37 · p. 281

Improper, quadratic factor ke saath

Nikaalo $\int \frac{x^{3}}{(x + 1) (x^{2} + 1)} d x .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 38NCERT Example 38 · p. 282

Parts jo cancel ho jaate hain

Nikaalo $\int \frac{\log x - 1}{(\log x)^{2}} d x,$ $x > 1.$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 39NCERT Example 39 · p. 283

Double substitution

Nikaalo $\int (\sqrt{\tan x} - \sqrt{\cot x}) d x,$ $0 < x < \frac{π}{2} .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 40NCERT Example 40 · p. 283

Chhupa hua square pehchaano

Nikaalo $\int \frac{\sin 2 x}{\sqrt{4 - \sin^{4} x}} d x .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 41NCERT Example 41 · p. 284

Product ka modulus

Evaluate karo $\int_{0}^{2} ∣ x \cos \frac{π x}{2} ∣ d x .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 42NCERT Example 42 · p. 284

Symmetry, ek substitution aur ek division

Evaluate karo $I = \int_{0}^{π} \frac{x \sin^{3} x}{1 + \cos^{2} x} d x .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Har integrand ko us method se match karo jo sabse pehle try karna chahiye.

Left mein koi item tap karo, phir right mein uska match.

Answer dekho

Answer

$x^{2} \cos (x^{3})$ : substitution
$\frac{x + 4}{(x - 1) (x + 2)}$ : partial fractions
$x^{2} \log x$ : integration by parts
$\sin^{2} 5 x$ : ek trigonometric identity

x^{2}

(ek-tihaai) derivative hai is function ka:

x^{3};

factorised denominator partial fractions ki taraf le jaata hai; power times log ke liye parts chahiye; squared sine is identity se simple hota hai:

\sin^{2} θ = \frac{1 - \cos 2 θ}{2} .

Recap

Summary aur key points

Integration differentiation ko ulta karta hai: $\int f (x) d x = F (x) + C$ jahan $F^{'} = f;$ ek interval par antiderivatives mein sirf ek constant ka farq hota hai (Section 7.2).
Linearity: $\int [k_{1} f_{1} + k_{2} f_{2}] d x = k_{1} \int f_{1} d x + k_{2} \int f_{2} d x$ (Section 7.2.1).
Har standard derivative ek standard integral deta hai, jaise $\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C$ ( $n \neq - 1$ ) aur $\int \frac{d x}{x} = \log ∣ x ∣ + C$ (Table 7.2).
Substitution: $t = g (x),$ lene par $\int f (g (x)) g^{'} (x) d x = \int f (t) d t .$ Isse milta hai $\int \tan x d x = \log ∣ \sec x ∣ + C$ aur $\int \sec x d x = \log ∣ \sec x + \tan x ∣ + C$ (Section 7.3.1).
$\cos^{2} x = \frac{1 + \cos 2 x}{2}$ jaisi identities aur product-to-sum formulas sines aur cosines ki powers aur products ko standard forms mein badal dete hain (Section 7.3.2).
Table 7.3 ke chhe special integrals handle karte hain $\frac{1}{x^{2} \pm a^{2}},$ $\frac{1}{a^{2} - x^{2}}$ aur unke square roots; completing the square kisi bhi quadratic ko inmein le aata hai (Section 7.4).
$\frac{p x + q}{quadratic}$ ke liye likho $p x + q = A \cdot (quadratic ka derivative) + B$ (Linear numerator).
Partial fractions: improper ho to pehle divide karo; har factor, yaani $x - a,$ $(x - a)^{2}$ ya irreducible $x^{2} + b x + c$ , Table 7.4 wale terms deta hai (Section 7.5).
Integration by parts: $\int f g d x = f \int g d x - \int (f^{'} \int g d x) d x;$ $f$ aisa chuno jo differentiate karne par simple ho jaaye (Section 7.6).
$\int e^{x} [f (x) + f^{'} (x)] d x = e^{x} f (x) + C$ (Section 7.6.1); teeno square-root integrals Table 7.5 mein hain.
$\int_{a}^{b} f (x) d x$ , $y = f (x)$ ke neeche $a$ se $b$ tak ka signed area hai (Section 7.7).
First fundamental theorem: $A (x) = \int_{a}^{x} f (t) d t$ ka derivative hai $A^{'} (x) = f (x) .$ Second: $\int_{a}^{b} f (x) d x = F (b) - F (a)$ , kisi bhi antiderivative $F$ ke liye (Section 7.8).
Definite integral mein substitution limits ko bhi badal deta hai (Section 7.9).
Properties P0–P7, khaas kar $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$ aur even/odd rule, jo is interval par lagta hai: $[- a, a],$ aksar bina antiderivative ke integral evaluate kar dete hain (Section 7.10).

Flash cards · 14

Jaldi yaad karo

1 / 14

Chapter 7 ki Miscellaneous Exercise

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercise 7.1, Q1–Q5	Practice 7.1 · Q1
NCERT Exercise 7.1, Q6–Q13	Practice 7.1 · Q2
NCERT Exercise 7.1, Q14–Q20	Practice 7.1 · Q3
NCERT Exercise 7.1, Q21	Practice 7.1 · Q4
NCERT Exercise 7.1, Q22	Practice 7.1 · Q5
NCERT Exercise 7.2, Q1–Q8	Practice 7.2 · Q1
NCERT Exercise 7.2, Q9–Q16	Practice 7.2 · Q2
NCERT Exercise 7.2, Q17–Q24	Practice 7.2 · Q3
NCERT Exercise 7.2, Q25–Q32	Practice 7.2 · Q4
NCERT Exercise 7.2, Q33–Q37	Practice 7.2 · Q5
NCERT Exercise 7.2, Q38	Practice 7.2 · Q6
NCERT Exercise 7.2, Q39	Practice 7.2 · Q7
NCERT Exercise 7.3, Q1–Q8	Practice 7.3 · Q1
NCERT Exercise 7.3, Q9–Q16	Practice 7.3 · Q2
NCERT Exercise 7.3, Q17–Q22	Practice 7.3 · Q3
NCERT Exercise 7.3, Q23	Practice 7.3 · Q4
NCERT Exercise 7.3, Q24	Practice 7.3 · Q5
NCERT Exercise 7.4, Q1–Q8	Practice 7.4 · Q1
NCERT Exercise 7.4, Q9–Q15	Practice 7.4 · Q2
NCERT Exercise 7.4, Q16–Q23	Practice 7.4 · Q3
NCERT Exercise 7.4, Q24	Practice 7.4 · Q4
NCERT Exercise 7.4, Q25	Practice 7.4 · Q5
NCERT Exercise 7.5, Q1–Q8	Practice 7.5 · Q1
NCERT Exercise 7.5, Q9–Q16	Practice 7.5 · Q2
NCERT Exercise 7.5, Q17–Q21	Practice 7.5 · Q3
NCERT Exercise 7.5, Q22	Practice 7.5 · Q4
NCERT Exercise 7.5, Q23	Practice 7.5 · Q5
NCERT Exercise 7.6, Q1–Q8	Practice 7.6 · Q1
NCERT Exercise 7.6, Q9–Q15	Practice 7.6 · Q2
NCERT Exercise 7.6, Q16–Q22	Practice 7.6 · Q3
NCERT Exercise 7.6, Q23	Practice 7.6 · Q4
NCERT Exercise 7.6, Q24	Practice 7.6 · Q5
NCERT Exercise 7.7, Q1–Q8	Practice 7.7 · Q1
NCERT Exercise 7.7, Q9	Practice 7.7 · Q2
NCERT Exercise 7.7, Q10	Practice 7.7 · Q3
NCERT Exercise 7.7, Q11	Practice 7.7 · Q4
NCERT Exercise 7.8, Q1–Q8	Practice 7.8 · Q1
NCERT Exercise 7.8, Q9–Q15	Practice 7.8 · Q2
NCERT Exercise 7.8, Q16–Q20	Practice 7.8 · Q3
NCERT Exercise 7.8, Q21	Practice 7.8 · Q4
NCERT Exercise 7.8, Q22	Practice 7.8 · Q5
NCERT Exercise 7.9, Q1–Q8	Practice 7.9 · Q1
NCERT Exercise 7.9, Q9	Practice 7.9 · Q2
NCERT Exercise 7.9, Q10	Practice 7.9 · Q3
NCERT Exercise 7.10, Q1–Q7	Practice 7.10 · Q1
NCERT Exercise 7.10, Q8–Q14	Practice 7.10 · Q2
NCERT Exercise 7.10, Q15–Q18	Practice 7.10 · Q3
NCERT Exercise 7.10, Q19	Practice 7.10 · Q4
NCERT Exercise 7.10, Q20	Practice 7.10 · Q5
NCERT Exercise 7.10, Q21	Practice 7.10 · Q6
NCERT Miscellaneous Exercise 7, Q1–Q8	Miscellaneous exercises · Q1
NCERT Miscellaneous Exercise 7, Q9–Q16	Miscellaneous exercises · Q2
NCERT Miscellaneous Exercise 7, Q17–Q23	Miscellaneous exercises · Q3
NCERT Miscellaneous Exercise 7, Q24–Q31	Miscellaneous exercises · Q4
NCERT Miscellaneous Exercise 7, Q32–Q37	Miscellaneous exercises · Q5
NCERT Miscellaneous Exercise 7, Q38	Miscellaneous exercises · Q6
NCERT Miscellaneous Exercise 7, Q39	Miscellaneous exercises · Q7
NCERT Miscellaneous Exercise 7, Q40	Miscellaneous exercises · Q8

Quick check: MCQs

7.2.1 Indefinite integral ki kuch properties4 questions
7.3.1 Integration by substitution4 questions
7.3.2 Trigonometric identities se integration3 questions
7.4 Kuch Khaas Functions ke Integrals4 questions
7.5 Partial Fractions se Integration3 questions
7.6.1 ∫eˣ[f(x) + f′(x)] dx type ka integral4 questions
7.6.2 Kuch aur types ke integrals2 questions
7.7 Definite Integral3 questions
7.8.3 Integral calculus ka second fundamental theorem4 questions
7.9 Substitution se Definite Integrals ka Evaluation3 questions
7.10 Definite Integrals ki Kuch Properties4 questions
Miscellaneous Examples3 questions

Glossary · 15

Key terms (English + Hindi)

antiderivativeप्रतिअवकलजprati-avakalaj: Aisa function $F$ jiska derivative kisi interval par $f$ ho; ise $f$ ka primitive bhi kehte hain.Chapter mein pehli baar yahan aaya →
area functionक्षेत्रफल फलनkshetraphal phalan: $A (x) = \int_{a}^{x} f (t) d t$ , graph ke neeche fixed $a$ se moving $x$ tak ka area.

completing the squareपूर्ण वर्ग बनानाpoorn varg banaana: $x^{2} + b x + c$ ko ${(x + \frac{b}{2})}^{2} + c - \frac{b^{2}}{4}$ ki tarah likhna, yaani ek square plus ek constant.Chapter mein pehli baar yahan aaya →
constant of integrationसमाकलन अचरsamaakalan achar: Antiderivative mein joda gaya arbitrary real number $C$ ; ek extra condition use fix kar deti hai.Chapter mein pehli baar yahan aaya →

definite integralनिश्चित समाकलनnishchit samaakalan: Number $\int_{a}^{b} f (x) d x$ : $y = f (x)$ aur $x$ -axis ke beech $x = a$ se $x = b$ tak ka signed area.Chapter mein pehli baar yahan aaya →

even and odd functionsसम और विषम फलनsam aur visham phalan: $f$ even hai agar $f (- x) = f (x)$ aur odd hai agar $f (- x) = - f (x)$ , apne domain ke har $x$ ke liye.Chapter mein pehli baar yahan aaya →

fundamental theorem of calculusकलन की आधारभूत प्रमेयkalan ki aadhaarbhoot prameya: Calculus ke do hisson ke beech ka link: $A^{'} (x) = f (x)$ , aur isliye $\int_{a}^{b} f (x) d x = F (b) - F (a)$ .Chapter mein pehli baar yahan aaya →

indefinite integralअनिश्चित समाकलनanishchit samaakalan: Kisi interval par $f$ ke saare antiderivatives ki family $\int f (x) d x = F (x) + C$ .Chapter mein pehli baar yahan aaya →
integrandसमाकल्यsamaakalya: Woh function $f (x)$ jo $\int f (x) d x$ ke andar hota hai.Chapter mein pehli baar yahan aaya →
integrationसमाकलनsamaakalan: Integral nikaalne ka process; indefinite integrals ke liye, antiderivatives nikaalna.Chapter mein pehli baar yahan aaya →
integration by partsखंडशः समाकलनkhandashah samaakalan: Rule $\int u v^{'} d x = u v - \int u^{'} v d x$ , yaani product rule ulta padha hua.Chapter mein pehli baar yahan aaya →
integration by substitutionप्रतिस्थापन द्वारा समाकलनpratisthaapan dvaara samaakalan: Variable badalna, $u = g (x)$ , taaki $\int f (g (x)) g^{'} (x) d x$ ban jaaye $\int f (u) d u$ .Chapter mein pehli baar yahan aaya →

limits of integrationसमाकलन की सीमाएँsamaakalan ki seemaaen: $\int_{a}^{b} f (x) d x$ ke end points $a$ (lower) aur $b$ (upper).Chapter mein pehli baar yahan aaya →

partial fractionsआंशिक भिन्नaanshik bhinn: Aise simple fractions jinka sum ek diye hue proper rational function ke barabar ho.Chapter mein pehli baar yahan aaya →

rational functionपरिमेय फलनparimeya phalan: Polynomials ka quotient $\frac{P (x)}{Q (x)}$ ; yeh proper hota hai jab $\deg P < \deg Q$ .Chapter mein pehli baar yahan aaya →

Integrals: Class 12 Maths Chapter 7 Hinglish mein

Is chapter mein aap kya seekhenge

7.1 Introduction

7.2 Integration: Differentiation ka Inverse Process

7.2.1 Indefinite integral ki kuch properties

Practice 7.1: answers ke saath

7.3 Integration ke Methods

7.3.1 Integration by substitution

Practice 7.2: answers ke saath

7.3.2 Trigonometric identities se integration

Practice 7.3: answers ke saath

7.4 Kuch Khaas Functions ke Integrals

Practice 7.4: answers ke saath

7.5 Partial Fractions se Integration

Practice 7.5: answers ke saath

7.6 Integration by Parts

7.6.1 ∫eˣ[f(x) + f′(x)] dx type ka integral

Practice 7.6: answers ke saath

7.6.2 Kuch aur types ke integrals

Practice 7.7: answers ke saath

7.7 Definite Integral

7.8 Fundamental Theorem of Calculus

7.8.1 Area function

7.8.2 Integral calculus ka first fundamental theorem

7.8.3 Integral calculus ka second fundamental theorem

Practice 7.8: answers ke saath

7.9 Substitution se Definite Integrals ka Evaluation

Practice 7.9: answers ke saath

7.10 Definite Integrals ki Kuch Properties

Practice 7.10: answers ke saath

Miscellaneous Examples

Summary aur key points

Chapter 7 ki Miscellaneous Exercise

Practice questions answers ke saath (NCERT “Miscellaneous exercises”)

Apna NCERT exercise question dhoondo

Quick check: MCQs

Key terms (English + Hindi)

Tum chapter ke end tak pahunch gaye