Integrals: Class 12 Maths · Chapter 7

Undoing the derivative, and measuring area

Mathematics Part 2 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

165 min
58 practice questions
41 MCQs

What you will learn in Integrals

Find antiderivatives by reading standard derivatives backwards, and fix the constant of integration from one given value
Use linearity, substitution and trigonometric identities to bring an integrand to a standard form
Integrate the special forms 1/(x² ± a²), 1/(a² − x²), 1/√(a² − x²), 1/√(x² ± a²) and the quadratic types that reduce to them by completing the square
Split a rational function into partial fractions and integrate it, dividing first when it is improper
Integrate products by parts, recognise the eˣ[f(x) + f′(x)] pattern, and integrate √(a² − x²) and √(x² ± a²)
Explain the definite integral as an area and use the area function and both fundamental theorems to evaluate definite integrals
Evaluate definite integrals by substitution with changed limits and by the properties P0–P7, including even and odd functions and the a + b − x reflection

Original companion text aligned to NCERT Mathematics Class 12, Chapter 7, pp. 225–291. Not affiliated with or endorsed by NCERT.

7.1 Introduction

A speedometer tells you how fast a bus is moving at each instant, but not how far it has gone. If you know the speed at every moment of a trip, can you work out the distance? Differentiation turns distance into speed. In this chapter we learn to go back the other way.

Integral calculus answers two kinds of question. The first runs differentiation backwards: given a function $f,$ find a function $F$ with $F^{'} = f .$ Such an $F$ is an antiderivative of $f .$ One formula covers every antiderivative at once; we call it the indefinite integral, and the search itself is integration (समाकलन). The second kind is about size: how much area lies between a graph and the $x$ -axis over an interval? That number is the definite integral.

3/3Problem 2: find the area between $y = f (x)$ and the $x$ -axis from $a$ to $b .$

Fig. 7.1The two problems that integral calculus answers.

Two side-by-side panels. In the first step the left panel shows only five short tangent segments at equally spaced x positions, each tilted by the known slope there; the right panel is faded. In the second step a smooth curve is drawn through the left panel so that it touches every slope mark: this is the function we were looking for. In the third step the right panel lights up: a rising curve above the x-axis with the region under it, between two marked points a and b, shaded.

At first sight these have nothing to do with each other. In Section 7.8 we will see that they share one answer: areas can be computed from antiderivatives. That result, the Fundamental Theorem of Calculus, replaces long limit calculations by a subtraction, and it is why integrals are everyday tools, from the work done by a varying force to the chance that a measurement falls in a range. This chapter first builds the toolkit for finding antiderivatives, then turns to definite integrals and the properties that make them easy to evaluate.

7.2 Integration as an Inverse Process of Differentiation

Differentiation takes a function and hands back its derivative. Integration runs the machine in reverse: start from the derivative and recover a function that produces it. Three facts you already know can be read backwards:

\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}

So $\tan x$ is an antiderivative of $\sec^{2} x,$ $\frac{x^{4}}{4}$ is one of $x^{3},$ and $- \cos x$ is one of $\sin x .$ But they are not the only ones. A constant has derivative zero, so $\frac{x^{4}}{4} + 7$ and $\frac{x^{4}}{4} - \sqrt{2}$ also have derivative $x^{3} .$ Adding any real number $C$ to an antiderivative gives another antiderivative. This $C$ is called the arbitrary constant or constant of integration; letting it run through all real numbers sweeps out a whole family of answers.

Definition

Antiderivative (primitive)

Let $f$ be defined on an interval $I .$ A function $F$ is an antiderivative (or primitive) of $f$ on $I$ if $F^{'} (x) = f (x)$ for every $x$ in $I .$ Then $F + C$ is also an antiderivative for every real constant $C .$

Example $F (x) = x^{2} - 5 x$ is an antiderivative of $f (x) = 2 x - 5$ on $𝐑,$ and so is $x^{2} - 5 x + 100.$

Graphically, the family $F (x) + C$ is one curve slid up and down. At any fixed $x$ every member has the same slope $f (x),$ so their tangents there are parallel (Fig. 7.2). One extra piece of information, such as a point the curve must pass through, picks out a single member.

2/3At $x = \frac{π}{3}$ every member has slope $\cos \frac{π}{3} = \frac{1}{2},$ so the tangents are parallel.

Fig. 7.2A family of antiderivatives of

f (x) = \cos x .

A grid with x from 0 to 2 pi. In the first step three curves of the form y equals sin x plus C are drawn for C equal to minus 3, 0 and 3, each a vertical shift of the same wave and each labelled with its value of C. In the second step a vertical dashed line at x equals pi by 3 crosses all three curves, and a short tangent segment at each crossing shows that all the tangents are parallel, with slope one half. In the third step the three curves fade and a new member, y equals sin x plus 1, is drawn through the marked point (pi by 2, 2).

TheoremAntiderivatives differ by a constant

If $g$ and $h$ have the same derivative at every point of an interval $I,$ then $g (x) - h (x)$ is a constant on $I .$

Let $ϕ (x) = g (x) - h (x) .$ Then $ϕ^{'} (x) = g^{'} (x) - h^{'} (x) = 0$ for every $x$ in $I .$ Take any two points $x_{1} < x_{2}$ of $I .$ By the mean value theorem there is a $c$ between them with $ϕ (x_{2}) - ϕ (x_{1}) = ϕ^{'} (c) (x_{2} - x_{1}) = 0.$ So $ϕ$ takes the same value everywhere on $I .$

The word interval matters. On the set $x \neq 0,$ which has two pieces, the function equal to $\log ∣ x ∣$ for $x > 0$ and $\log ∣ x ∣ + 5$ for $x < 0$ still has derivative $\frac{1}{x},$ yet it is not $\log ∣ x ∣$ plus one constant. Within any single interval, though, the family ${F + C}$ catches every antiderivative.

Because of this, one symbol can stand for the whole family. We write $\int f (x) d x,$ read “the integral of $f$ with respect to $x$ ”, and if $F$ is any one antiderivative,

Indefinite integral

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

In the same spirit, if $\frac{d y}{d x} = f (x)$ we write $y = \int f (x) d x .$ The words that go with the symbol are collected in Table 7.1.

Table 7.1The language of indefinite integralsNCERT Table 7.1 · p. 227

Symbol or word	What it means
$\int f (x) d x$	the whole family of antiderivatives of $f,$ with $x$ as the variable
$f (x)$	the integrand, the function being integrated
$d x$	says that $x$ is the variable of integration
integrate	find the indefinite integral
an integral of $f$	any one function $F$ with $F^{'} (x) = f (x)$
integration	the reverse of differentiation: working out integrals
$C$	the constant of integration, any real number

Definition

Indefinite integral

The indefinite integral $\int f (x) d x$ is the family of all antiderivatives of $f$ on an interval, written $F (x) + C$ where $F$ is any one antiderivative and $C$ is an arbitrary constant.

Every derivative formula from your differentiation toolkit turns into an integral formula when read from right to left. Table 7.2 lists the ones we use constantly. You can check each row by differentiating its right-hand side.

Table 7.2Standard integrals: differentiating the right-hand entry gives the left-hand oneNCERT Standard formulae · p. 228

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

The last two rows offer two answers each. They do not contradict each other: $\sin^{- 1} x$ and $- \cos^{- 1} x$ differ by the constant $\frac{π}{2},$ so both are correct. Each formula also holds only where both sides are defined. $\int \frac{1}{x} d x = \log ∣ x ∣ + C$ is valid on any interval that avoids $0,$ and a formula with a square root needs the quantity under the root to be positive. These intervals are rarely written out, so make a habit of checking them yourself.

7.2.1 Some properties of indefinite integral

The rules below let us break a complicated integrand into pieces we can read off Table 7.2.

PropertyIntegration and differentiation undo each other

(I) Differentiating an indefinite integral gives back the integrand, $\frac{d}{d x} \int u d x = u;$ integrating a derivative gives back the function up to a constant, $\int u^{'} d x = u + C .$ (II) If two indefinite integrals have the same derivative, they describe the same family of curves; we then write them as equal.

(I) If $\int u d x = U + C$ with $U^{'} = u,$ then $\frac{d}{d x} (U + C) = U^{'} = u .$ For the second half, $u$ is itself a function whose derivative is $u^{'},$ so it is one antiderivative of $u^{'},$ and the family is $u + C .$

(II) Suppose $\frac{d}{d x} \int f d x = \frac{d}{d x} \int g d x .$ The difference of the two integrals then has derivative $0$ on the interval, so by the theorem above it is a constant. Each family is the other shifted by a constant, which is the same family. That is the sense in which an equation such as $\int f d x = \int g d x$ is read: equal up to the arbitrary constant.

PropertyLinearity

(III) The integral of a sum is the sum of the integrals: $\int (u + v) d x = \int u d x + \int v d x .$ (IV) A constant factor can move outside: $\int k u d x = k \int u d x$ for every real $k .$ (V) Combining the two, for functions $u_{1}, \dots, u_{n}$ and constants $k_{1}, \dots, k_{n},$ the integral of $k_{1} u_{1} + \dots + k_{n} u_{n}$ is $k_{1} \int u_{1} d x + \dots + k_{n} \int u_{n} d x .$

With these rules the simplest way to integrate is the method of inspection: guess a function whose derivative is the integrand, then differentiate your guess to confirm it. Adjust a constant factor if the check is off by one.

Example 1NCERT Example 1 · p. 231

Antiderivatives by inspection

Write down an antiderivative of each function: (i) $\sin 3 x$ (ii) $5 x^{4} - 6 x^{2}$ (iii) $\frac{3}{x},$ $x \neq 0.$

Try it yourself first. Then reveal the solution one step at a time.

Example 2NCERT Example 2 · p. 232

Using linearity

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

Try it yourself first. Then reveal the solution one step at a time.

Example 3NCERT Example 3 · p. 233

Trigonometric integrands

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

Try it yourself first. Then reveal the solution one step at a time.

Example 4NCERT Example 4 · p. 233

From velocity to position

A toy train moves along a straight track. Its velocity after $t$ seconds is $v (t) = 3 t^{2} + 2 t$ (in $c m / s$ ), and at $t = 0$ it is $5 c m$ from the start of the track. Find its position $s (t),$ and where it is at $t = 2.$

Try it yourself first. Then reveal the solution one step at a time.

Example 4 shows the general pattern. The family $F + C$ contains infinitely many antiderivatives; one extra condition, such as a starting value, selects exactly one of them. The variable need not be $x$ : $\int t^{6} d t = \frac{t^{7}}{7} + C$ and $\int u^{- 3} d u = - \frac{1}{2 u^{2}} + C$ follow the same rules.

Practice · 5 questions

Practice 7.1: questions with answers

1
Short answerEasyNCERT Exercise 7.1, Q1–Q5
Guess an antiderivative of each function, then check it by differentiating.
1. $\cos 4 x$
  Show solution
  Answer $\frac{1}{4} \sin 4 x$
  $\frac{d}{d x} \sin 4 x = 4 \cos 4 x,$ so divide by $4$ : $\frac{d}{d x} (\frac{1}{4} \sin 4 x) = \cos 4 x .$
2. $\sin 5 x$
  Show solution
  Answer $- \frac{1}{5} \cos 5 x$
  $\frac{d}{d x} \cos 5 x = - 5 \sin 5 x,$ so $- \frac{1}{5} \cos 5 x$ has derivative $\sin 5 x .$
2
Short answerEasyNCERT Exercise 7.1, Q6–Q13
Find each integral.
1. $\int (2 e^{2 x} - 3) d x$
  Show solution
  Answer $e^{2 x} - 3 x + C$
  $\int 2 e^{2 x} d x = e^{2 x}$ (check: derivative $2 e^{2 x}$ ) and $\int 3 d x = 3 x .$
2. $\int x^{3} (1 + \frac{2}{x^{2}}) d x$
  Show solution
  Answer $\frac{x^{4}}{4} + x^{2} + C$
  Expand: $x^{3} + 2 x .$ Then $\int (x^{3} + 2 x) d x = \frac{x^{4}}{4} + x^{2} + C .$
3
Short answerMediumNCERT Exercise 7.1, Q14–Q20
Find each integral.
1. $\int (2 + x) \sqrt{x} d x$
  Show solution
  Answer $\frac{4}{3} x^{3 / 2} + \frac{2}{5} x^{5 / 2} + C$
  Expand: $2 x^{1 / 2} + x^{3 / 2},$ giving $2 \cdot \frac{2}{3} x^{3 / 2} + \frac{2}{5} x^{5 / 2} + C .$
2. $\int \sqrt{x} (2 x^{2} - x + 4) d x$
  Show solution
  Answer $\frac{4}{7} x^{7 / 2} - \frac{2}{5} x^{5 / 2} + \frac{8}{3} x^{3 / 2} + C$
  Expand: $2 x^{5 / 2} - x^{3 / 2} + 4 x^{1 / 2} .$
  Integrate: $2 \cdot \frac{2}{7} x^{7 / 2} - \frac{2}{5} x^{5 / 2} + 4 \cdot \frac{2}{3} x^{3 / 2} + C .$
Quick answerEasyNCERT Exercise 7.1, Q21
An antiderivative of $2 \sqrt{x} - \frac{3}{\sqrt{x}}$ is
$\frac{4}{3} x^{3 / 2} - \frac{3}{2} x^{1 / 2} + C$ $x^{- 1 / 2} + \frac{3}{2} x^{- 3 / 2} + C$ $\frac{4}{3} x^{3 / 2} - 6 x^{1 / 2} + C$ $3 x^{3 / 2} - 6 x^{1 / 2} + C$
Show answer
Answer (C) $\frac{4}{3} x^{3 / 2} - 6 x^{1 / 2} + C$
$\int 2 x^{1 / 2} d x = \frac{4}{3} x^{3 / 2}$ and $\int 3 x^{- 1 / 2} d x = 6 x^{1 / 2} .$ Option B differentiates instead of integrating; option A multiplies by the exponent instead of dividing by it.
Quick answerMediumNCERT Exercise 7.1, Q22
If $\frac{d}{d x} f (x) = 6 x^{2} - \frac{2}{x^{3}}$ and $f (1) = 4,$ then $f (x)$ is
$2 x^{3} - \frac{1}{x^{2}} + 3$ $2 x^{3} + \frac{1}{x^{2}} + 1$ $2 x^{3} + \frac{1}{x^{2}} + 4$ $12 x + \frac{6}{x^{4}} - 14$
Show answer
Answer (B) $2 x^{3} + \frac{1}{x^{2}} + 1$
$\int (6 x^{2} - 2 x^{- 3}) d x = 2 x^{3} + x^{- 2} + C,$ and $f (1) = 3 + C = 4$ gives $C = 1.$ Option A gets the sign of $\int - 2 x^{- 3} d x$ wrong; option C forgets that $f (1)$ already includes $2 + 1.$

7.3 Methods of Integration

Inspection only works when we recognise the integrand as a derivative, and most integrands are not that obliging. So we need ways to transform an integral until it matches a row of Table 7.2. Three are central, and each comes from a rule of differentiation: substitution undoes the chain rule, partial fractions undo the adding of fractions, and integration by parts undoes the product rule.

7.3.1 Integration by substitution

The chain rule says that if $u = g (x)$ then $\frac{d}{d x} F (g (x)) = F^{'} (g (x)) g^{'} (x) .$ Read it backwards: an integrand of the form (function of $g (x)$ ) × $g^{'} (x)$ is the derivative of something simple. Fig. 7.3 shows how to spot this.

2/3In $\int \cos (x^{2}) 2 x d x$ look for an inner function whose derivative is also present.

Fig. 7.3Substitution is the chain rule read backwards.

In the first step a pipeline of three boxes reads x, then u equals x squared, then sin u, with an arrow labelled differentiate leading to cos of x squared times 2x. In the second step the integrand cos of x squared times 2x is shown with a solid box around cos of x squared, labelled function of x squared, and a dashed box around the factor 2x, labelled derivative of x squared. In the third step the dashed factor and dx merge into du, the integral becomes the integral of cos u du, and the result sin of x squared plus C appears.

In symbols: if we write $x = ϕ (u),$ then $d x = ϕ^{'} (u) d u,$ and the whole integral can be rewritten in terms of $u$ :

Change of variable

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

In practice we usually go the other way and name a piece of the integrand: $t = h (x),$ $d t = h^{'} (x) d x .$ Choosing $t$ is the skill to develop. Look for a piece of the integrand whose derivative is also present, perhaps up to a constant factor; that piece is the natural choice.

Step by step

Working rule for substitution

Step 1 of 4

Choose

Pick an inner expression $t = h (x)$ whose derivative $h^{'} (x)$ appears (possibly times a constant) in the integrand.

1 / 4

Example 5NCERT Example 5 · p. 236

Four substitutions

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

Try it yourself first. Then reveal the solution one step at a time.

In (i) the inner function was linear. That case is worth remembering on its own: if $\int f (x) d x = F (x) + C,$ then $\int f (a x + b) d x = \frac{1}{a} F (a x + b) + C$ for any $a \neq 0.$

Substitution also handles the four trigonometric functions missing from our table. Learn the results below: they turn up constantly.

DerivationIntegrals of tan, cot, sec and cosec

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

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

plus a constant. Since $- \log ∣ \cos x ∣ = \log ∣ \sec x ∣,$ this is $\int \tan x d x = \log ∣ \sec x ∣ + C .$

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

sec. Multiply above and below by $\sec x + \tan x .$ The new numerator $\sec^{2} x + \sec x \tan x$ is exactly the derivative of the new denominator, so with $t = \sec x + \tan x$ :

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

That is, $\int \sec x d x = \log ∣ \sec x + \tan x ∣ + C .$

cosec. Multiply above and below by $c o s e c x - \cot x .$ The numerator becomes ${c o s e c}^{2} x - c o s e c x \cot x,$ which is the derivative of $t = c o s e c x - \cot x$ (check: $\frac{d}{d x} c o s e c x = - c o s e c x \cot x$ and $\frac{d}{d x} (- \cot x) = {c o s e c}^{2} x$ ). So again:

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

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

Since $(c o s e c x - \cot x) (c o s e c x + \cot x) = 1,$ the same answer can be written $- \log ∣ c o s e c x + \cot x ∣ + C .$

Example 6NCERT Example 6 · p. 238

Substitution with a little preparation

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

Try it yourself first. Then reveal the solution one step at a time.

Practice · 7 questions

Practice 7.2: questions with answers

1
Short answerEasyNCERT Exercise 7.2, Q1–Q8
Integrate each function with a suitable substitution.
1. $\frac{3 x^{2}}{4 + x^{3}}$
  Show solution
  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. $\frac{(\log x)^{4}}{x}$
  Show solution
  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
Integrate each function.
1. $(6 x - 3) \sqrt{x^{2} - x + 2}$
  Show solution
  Answer $2 (x^{2} - x + 2)^{3 / 2} + C$
  $t = x^{2} - x + 2,$ $d t = (2 x - 1) d x,$ and $6 x - 3 = 3 (2 x - 1)$ : $3 \int t^{1 / 2} d t = 2 t^{3 / 2} + C .$
2. $\frac{1}{x + \sqrt{x}},$ $x > 0$
  Show solution
  Answer $2 \log (\sqrt{x} + 1) + C$
  Factor: $\frac{1}{\sqrt{x} (\sqrt{x} + 1)} .$ Put $t = \sqrt{x} + 1,$ $d t = \frac{d x}{2 \sqrt{x}}$ : $2 \int \frac{d t}{t} = 2 \log t + C .$
3
Short answerMediumNCERT Exercise 7.2, Q17–Q24
Integrate each function.
1. $x^{2} e^{- x^{3}}$
  Show solution
  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. $\frac{(\tan^{- 1} x)^{2}}{1 + x^{2}}$
  Show solution
  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
Integrate each function.
1. $\frac{\sec^{2} x}{(2 + \tan x)^{3}}$
  Show solution
  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. $\frac{\sin \sqrt{x}}{\sqrt{x}}$
  Show solution
  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
Integrate each function.
1. $\frac{\cos x}{\cos x + 3 \sin x}$
  Show solution
  Answer $\frac{x}{10} + \frac{3}{10} \log ∣ \cos x + 3 \sin x ∣ + C$
  Write $\cos x = A (\cos x + 3 \sin x) + B (3 \cos x - \sin x),$ the second bracket being the derivative of the denominator.
  Matching $\cos x$ : $A + 3 B = 1;$ matching $\sin x$ : $3 A - B = 0.$ So $A = \frac{1}{10},$ $B = \frac{3}{10},$ and the integral is $A x + B \log ∣ \cos x + 3 \sin x ∣ + C .$
2. $\frac{\sqrt{\cot x}}{\sin x \cos x},$ $0 < x < \frac{π}{2}$
  Show solution
  Answer $- 2 \sqrt{\cot x} + C$
  $\frac{1}{\sin x \cos x} = \frac{{c o s e c}^{2} x}{\cot x},$ so the integrand is $(\cot x)^{- 1 / 2} {c o s e c}^{2} x .$
  $t = \cot x,$ $d t = - {c o s e c}^{2} x d x$ : $- \int t^{- 1 / 2} d t = - 2 \sqrt{t} .$
Quick answerEasyNCERT Exercise 7.2, Q38
$\int \frac{3 x^{2} + 2^{x} \log 2}{x^{3} + 2^{x}} d x$ equals
$(x^{3} + 2^{x})^{- 1} + C$ $\log ∣ 3 x^{2} + 2^{x} \log 2 ∣ + C$ $x^{3} + 2^{x} + C$ $\log ∣ x^{3} + 2^{x} ∣ + C$
Show answer
Answer (D) $\log ∣ x^{3} + 2^{x} ∣ + C$
The numerator is exactly the derivative of $x^{3} + 2^{x},$ so the integral is $\log ∣ x^{3} + 2^{x} ∣ + C .$ Option B takes the log of the wrong expression.
Quick answerMediumNCERT Exercise 7.2, Q39
$\int \frac{4 d x}{\sin^{2} 2 x}$ equals
$- \cot 2 x + C$ $\tan x - \cot x + C$ $4 \cot 2 x + C$ $\tan x + \cot x + C$
Show answer
Answer (B) $\tan x - \cot x + C$
$\frac{4}{\sin^{2} 2 x} = \frac{1}{\sin^{2} x \cos^{2} x} = \frac{\sin^{2} x + \cos^{2} x}{\sin^{2} x \cos^{2} x} = \sec^{2} x + {c o s e c}^{2} x,$ which integrates to $\tan x - \cot x .$ (This also equals $- 2 \cot 2 x,$ so option A is off by a factor of 2.)

7.3.2 Integration using trigonometric identities

Squares, cubes and products of sines and cosines are not in our table, but identities rewrite them as sums of single sines and cosines, which are. The three workhorses are the double-angle formulas in the form $\cos^{2} x = \frac{1 + \cos 2 x}{2}$ and $\sin^{2} x = \frac{1 - \cos 2 x}{2},$ the product-to-sum formulas such as $2 \cos A \cos B = \cos (A + B) + \cos (A - B),$ and the triple-angle formulas. Fig. 7.4 shows why the first one is true.

3/3It is a wave of height $\frac{1}{2}$ about the line $y = \frac{1}{2}$ : $\cos^{2} x = \frac{1}{2} + \frac{1}{2} \cos 2 x .$

Fig. 7.4Squaring a cosine gives a faster, smaller cosine on top of a constant.

A graph for x from 0 to 2 pi. In the first step the cosine curve swings between minus 1 and 1. In the second step the curve cos squared x is added: it never goes below 0, touches 1 at 0, pi and 2 pi, touches 0 at pi by 2 and 3 pi by 2, and repeats twice as often. In the third step a dashed horizontal line at height one half appears and the cos squared curve is seen to be a cosine wave of amplitude one half and period pi riding on that line, which is the identity cos squared x equals one half plus one half cos 2x.

Example 7NCERT Example 7 · p. 241

Three identities in action

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

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

Which identity turns

\sin^{2} 4 x

into something we can integrate directly?

\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}

Show answer

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

The double-angle form removes the square:

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

integrates term by term. Option C uses the wrong angle (it should double

4 x

8 x

); option A just swaps one square for another.

Practice · 5 questions

Practice 7.3: questions with answers

1
Short answerMediumNCERT Exercise 7.3, Q1–Q8
Integrate each function using an identity.
1. $\cos^{2} (3 x - 1)$
  Show solution
  Answer $\frac{x}{2} + \frac{1}{12} \sin (6 x - 2) + C$
  $\cos^{2} θ = \frac{1 + \cos 2 θ}{2}$ with $θ = 3 x - 1$ : $\int \frac{1 + \cos (6 x - 2)}{2} d x .$
2. $\sin 5 x \cos 2 x$
  Show solution
  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
Integrate each function.
1. $\frac{\cos x}{1 - \cos x}$
  Show solution
  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. $\cos^{4} x$
  Show solution
  Answer $\frac{3 x}{8} + \frac{\sin 2 x}{4} + \frac{\sin 4 x}{32} + C$
  $\cos^{4} x = (\frac{1 + \cos 2 x}{2})^{2} = \frac{1}{4} (1 + 2 \cos 2 x + \cos^{2} 2 x),$ and $\cos^{2} 2 x = \frac{1 + \cos 4 x}{2} .$
  So $\cos^{4} x = \frac{3}{8} + \frac{1}{2} \cos 2 x + \frac{1}{8} \cos 4 x .$
3
Short answerHardNCERT Exercise 7.3, Q17–Q22
Integrate each function.
1. $\frac{\sin^{3} x - \cos^{3} x}{\sin^{2} x \cos^{2} x}$
  Show solution
  Answer $\sec x + c o s e c x + C$
  Split: $\frac{\sin x}{\cos^{2} x} - \frac{\cos x}{\sin^{2} x} = \sec x \tan x - c o s e c x \cot x .$
2. $\frac{\cos 2 x - 2 \cos^{2} x}{\sin^{2} x}$
  Show solution
  Answer $\cot x + C$
  $\cos 2 x - 2 \cos^{2} x = - 1,$ so the integrand is $- {c o s e c}^{2} x .$
Quick answerMediumNCERT Exercise 7.3, Q23
$\int (\tan x - \cot x)^{2} d x$ equals
$\tan x - \cot x - 4 x + C$ $\tan x + \cot x - 2 x + C$ $\frac{1}{3} (\tan x - \cot x)^{3} + C$ $\sec^{2} x + {c o s e c}^{2} x - 4 x + C$
Show answer
Answer (A) $\tan x - \cot x - 4 x + C$
$(\tan x - \cot x)^{2} = \tan^{2} x + \cot^{2} x - 2 = \sec^{2} x + {c o s e c}^{2} x - 4,$ which integrates to $\tan x - \cot x - 4 x .$ Option C treats the bracket like $x$ in the power rule, which is wrong because its derivative is not 1.
Quick answerMediumNCERT Exercise 7.3, Q24
$\int (1 + \log x) \sec^{2} (x \log x) d x$ equals
$\sec (x \log x) + C$ $x \tan (\log x) + C$ $\log ∣ \tan x ∣ + C$ $\tan (x \log x) + C$
Show answer
Answer (D) $\tan (x \log x) + C$
$\frac{d}{d x} (x \log x) = 1 + \log x,$ so with $t = x \log x$ the integral is $\int \sec^{2} t d t = \tan t + C .$

7.4 Integrals of Some Particular Functions

Six integrands with $x^{2}$ and a constant $a^{2}$ (where $a > 0$ ) appear so often that we prove them once and then quote them. Together with completing the square they handle every integrand of the form $\frac{1}{quadratic}$ or $\frac{1}{\sqrt{quadratic}} .$

Table 7.3Six special integrals (the numbering is used throughout the chapter)NCERT 7.4 (1)–(6) · p. 243

	Integrand	Integral (add $C$ )
(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)	$\frac{1}{\sqrt{x^{2} + a^{2}}}$	$\log ∣ x + \sqrt{x^{2} + a^{2}} ∣$

DerivationWhy the six formulas hold

(1) and (2): split the fraction. Since $(x + a) - (x - a) = 2 a,$

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

Integrating gives $\frac{1}{2 a} (\log ∣ x - a ∣ - \log ∣ x + a ∣),$ which is (1). In the same way $\frac{1}{a^{2} - x^{2}} = \frac{1}{2 a} (\frac{1}{a - x} + \frac{1}{a + x});$ since $\int \frac{d x}{a - x} = - \log ∣ a - x ∣,$ we get (2). Splitting a fraction like this is the subject of Section 7.5.

(3) and (5): a trigonometric substitution. For (3) put $x = a \tan θ,$ so $d x = a \sec^{2} θ d θ$ and $x^{2} + a^{2} = a^{2} \sec^{2} θ$ :

\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}

For (5) put $x = a \sin θ;$ then $\sqrt{a^{2} - x^{2}} = a \cos θ$ and $d x = a \cos θ d θ,$ so the integral is $\int d θ = θ = \sin^{- 1} \frac{x}{a} .$

(4) and (6): check by differentiating. Let $s = \sqrt{x^{2} \pm a^{2}},$ so $\frac{d s}{d x} = \frac{x}{s} .$ Then

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

which is exactly the integrand. (The same answers also come from the substitutions $x = a \sec θ$ and $x = a \tan θ,$ with $\int \sec θ d θ$ from Section 7.3.1; that route produces an extra $- \log a,$ which is a constant and is absorbed into $C$ .)

Quadratics in general. An integrand $\frac{1}{a x^{2} + b x + c}$ or $\frac{1}{\sqrt{a x^{2} + b x + c}}$ is brought to one of the six forms by completing the square. Factor out $a,$ then add and subtract the square of half the coefficient of $x$ :

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}

With $t = x + \frac{b}{2 a}$ (so $d t = d x$ ) the bracket becomes $t^{2} + k^{2}$ or $t^{2} - k^{2},$ depending on the sign of $\frac{c}{a} - \frac{b^{2}}{4 a^{2}} .$ Fig. 7.5 shows the idea with tiles.

3/4The missing corner has area $2^{2} = 4,$ so $x^{2} + 4 x + 4 = (x + 2)^{2} .$

Fig. 7.5Completing the square, drawn with tiles.

In the first step a large square of side x, labelled x squared, sits next to a long rectangle of width 4 and height x, labelled 4x. In the second step the rectangle is cut into two strips of width 2: one is moved under the square and turned, so the square has a strip along its right side and one along its bottom, leaving an empty 2 by 2 corner drawn dashed. In the third step a small 2 by 2 square fills the corner, completing a square of side x plus 2. In the fourth step the text beside the square reads x squared plus 4x plus 20 equals x plus 2, all squared, plus 16.

Linear over quadratic. For $\int \frac{p x + q}{a x^{2} + b x + c} d x$ and $\int \frac{p x + q}{\sqrt{a x^{2} + b x + c}} d x$ we split the numerator into a multiple of the derivative of the quadratic plus a constant:

Linear numerator

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

Step by step

Working rule for linear over quadratic

Step 1 of 4

Match coefficients

Compare the coefficient of $x$ ( $p = 2 a A$ ) and the constant term ( $q = A b + B$ ) to find $A$ and $B .$

1 / 4

Example 8NCERT Example 8 · p. 247

Straight from the table

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

Try it yourself first. Then reveal the solution one step at a time.

Example 9NCERT Example 9 · p. 247

Complete the square first

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

Try it yourself first. Then reveal the solution one step at a time.

Example 10NCERT Example 10 · p. 249

A linear numerator

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

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 4

Match each integrand with the formula that finishes it.

Tap an item on the left, then its match on the right.

Show answer

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 and no root: inverse tangent. Root of

a^{2} - x^{2}

: inverse sine. Minus sign, no root: logarithm of a ratio. Root of

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

: logarithm of

x

plus the root.

Practice · 5 questions

Practice 7.4: questions with answers

1
Short answerEasyNCERT Exercise 7.4, Q1–Q8
Integrate each function using formulas (1)–(6).
1. $\frac{4 x^{3}}{x^{8} + 9}$
  Show solution
  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).
2. $\frac{1}{\sqrt{9 x^{2} + 4}}$
  Show solution
  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).
2
Short answerMediumNCERT Exercise 7.4, Q9–Q15
Complete the square, then integrate.
1. $\frac{\cos x}{\sqrt{\sin^{2} x + 9}}$
  Show solution
  Answer $\log ∣ \sin x + \sqrt{\sin^{2} x + 9} ∣ + C$
  $t = \sin x$ : $\int \frac{d t}{\sqrt{t^{2} + 3^{2}}},$ formula (6).
2. $\frac{1}{\sqrt{x^{2} - 6 x + 13}}$
  Show solution
  Answer $\log ∣ x - 3 + \sqrt{x^{2} - 6 x + 13} ∣ + C$
  $x^{2} - 6 x + 13 = (x - 3)^{2} + 2^{2};$ formula (6) with $t = x - 3.$
3
Short answerHardNCERT Exercise 7.4, Q16–Q23
Integrate each function (linear numerator).
1. $\frac{6 x - 1}{\sqrt{3 x^{2} - x + 2}}$
  Show solution
  Answer $2 \sqrt{3 x^{2} - x + 2} + C$
  The numerator is exactly the derivative of $3 x^{2} - x + 2,$ so with $t = 3 x^{2} - x + 2$ the integral is $\int t^{- 1 / 2} d t = 2 \sqrt{t} .$
2. $\frac{x - 3}{\sqrt{x^{2} + 4}}$
  Show solution
  Answer $\sqrt{x^{2} + 4} - 3 \log ∣ x + \sqrt{x^{2} + 4} ∣ + C$
  $\int \frac{x d x}{\sqrt{x^{2} + 4}} = \sqrt{x^{2} + 4},$ and $3 \int \frac{d x}{\sqrt{x^{2} + 4}}$ is formula (6).
Quick answerEasyNCERT Exercise 7.4, Q24
$\int \frac{d x}{x^{2} - 4 x + 5}$ equals
$x \tan^{- 1} (x - 2) + C$ $(x - 2) \tan^{- 1} x + C$ $\tan^{- 1} (x - 2) + C$ $\tan^{- 1} x + C$
Show answer
Answer (C) $\tan^{- 1} (x - 2) + C$
$x^{2} - 4 x + 5 = (x - 2)^{2} + 1,$ and formula (3) with $a = 1$ gives $\tan^{- 1} (x - 2) .$
Quick answerMediumNCERT Exercise 7.4, Q25
$\int \frac{d x}{\sqrt{5 x - 9 x^{2}}}$ equals
$\frac{1}{9} \sin^{- 1} \frac{18 x - 5}{5} + C$ $\sin^{- 1} \frac{18 x - 5}{5} + C$ $\frac{1}{3} \sin^{- 1} \frac{18 x - 5}{5} + C$ $\frac{1}{3} \sin^{- 1} \frac{9 x - 5}{5} + C$
Show answer
Answer (C) $\frac{1}{3} \sin^{- 1} \frac{18 x - 5}{5} + C$
$5 x - 9 x^{2} = 9 [(\frac{5}{18})^{2} - (x - \frac{5}{18})^{2}],$ so the integral is $\frac{1}{3} \sin^{- 1} \frac{x - 5 / 18}{5 / 18} = \frac{1}{3} \sin^{- 1} \frac{18 x - 5}{5} .$ Option A uses $\frac{1}{9}$ instead of $\frac{1}{\sqrt{9}} .$

7.5 Integration by Partial Fractions

A rational function is a quotient $\frac{P (x)}{Q (x)}$ of two polynomials with $Q (x) \neq 0.$ Adding fractions is easy: $\frac{2}{x - 1} + \frac{1}{x + 2} = \frac{3 x + 3}{(x - 1) (x + 2)} .$ Integration needs the reverse. Given the single fraction on the right, we recover the simple pieces on the left, called partial fractions (आंशिक भिन्न). Each piece integrates to a logarithm, a power or an inverse tangent.

Definition

Proper and improper rational functions

Call $\frac{P (x)}{Q (x)}$ proper when the top has lower degree than the bottom ( $\deg P < \deg Q$ ), and improper when $\deg P \geq \deg Q .$ Dividing $P$ by $Q$ writes an improper fraction as a polynomial $T (x)$ plus a proper fraction $\frac{P_{1} (x)}{Q (x)} .$

Example $\frac{x^{2} + 2}{x^{2} - 3 x + 2}$ is improper (degrees equal) and equals $1 + \frac{3 x}{x^{2} - 3 x + 2} .$

The polynomial part left over after division integrates directly, so the real work is with proper fractions. Table 7.4 covers every denominator we will meet: products of linear factors, possibly repeated, and quadratics with no real roots. Each factor contributes terms with unknown constants $A, B, C, \dots;$ the partial fractions of the whole are the sum of these contributions.

Table 7.4Which partial fractions to write down for a proper fractionNCERT Table 7.2 · p. 253

Factor of $Q (x)$	Terms it adds
$x - a$	$\frac{A}{x - a}$
$(x - a)^{2}$	$\frac{A}{x - a} + \frac{B}{(x - a)^{2}}$
$x^{2} + b x + c$ with no real roots	$\frac{B x + C}{x^{2} + b x + c}$

For example, $\frac{x + 5}{(x - 1)^{2} (x + 2)}$ is written $\frac{A}{x - 1} + \frac{B}{(x - 1)^{2}} + \frac{C}{x + 2},$ and $\frac{3 x^{2} + 2}{(x - 1) (x^{2} + 4)}$ is written $\frac{A}{x - 1} + \frac{B x + C}{x^{2} + 4} .$ To find the constants, multiply through by $Q (x)$ and compare coefficients, or substitute convenient values of $x .$ Fig. 7.6 shows what the splitting does to a graph.

3/3Added together: $\frac{1}{x - 1} - \frac{1}{x + 1} = \frac{2}{x^{2} - 1},$ with both behaviours.

Fig. 7.6A rational function as the sum of two simple pieces.

A coordinate grid with x from minus 4 to 4. In the first step the graph of y equals 1 over x minus 1 is drawn, with a dashed vertical asymptote at x equals 1. In the second step the graph of y equals minus 1 over x plus 1 is added, with a dashed asymptote at x equals minus 1. In the third step both fade and their sum, y equals 2 over x squared minus 1, is drawn in a thick line: it has both asymptotes, dips to minus 2 at x equals 0 between them, and approaches 0 far away on both sides.

Example 11NCERT Example 11 · p. 254

Two distinct linear factors

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

Try it yourself first. Then reveal the solution one step at a time.

The line $1 = A (x + 3) + B (x - 1)$ in Example 11 is an identity: true for every $x,$ not just for some (some books write $\equiv$ for this). That is what allows a faster route: put $x = 1$ to get $1 = 4 A,$ and $x = - 3$ to get $1 = - 4 B .$ Substituting the roots of the factors kills all but one unknown at a time.

Example 12NCERT Example 12 · p. 254

An improper fraction: divide first

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

Try it yourself first. Then reveal the solution one step at a time.

Example 13NCERT Example 13 · p. 255

A repeated linear factor

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

Try it yourself first. Then reveal the solution one step at a time.

Example 14NCERT Example 14 · p. 256

Treat x² as one variable

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

Try it yourself first. Then reveal the solution one step at a time.

In Example 14 the $y$ was a device for the algebra, not a substitution in the integral. The next example uses a genuine substitution first, and only then partial fractions.

Example 15NCERT Example 15 · p. 256

Substitute, then split

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

Try it yourself first. Then reveal the solution one step at a time.

Example 16NCERT Example 16 · p. 257

A quadratic factor

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

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

Which is the correct form to start with for

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

\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}

Show answer

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

An irreducible quadratic factor needs a linear numerator

B x + C .

Option A has too few unknowns to match every numerator, and option D factors

x^{2} + 1,

which has no real roots.

Practice · 5 questions

Practice 7.5: questions with answers

1
Short answerEasyNCERT Exercise 7.5, Q1–Q8
Split into partial fractions and integrate.
1. $\frac{x}{(x - 2) (x + 3)}$
  Show solution
  Answer $\frac{2}{5} \log ∣ x - 2 ∣ + \frac{3}{5} \log ∣ x + 3 ∣ + C$
  $x = A (x + 3) + B (x - 2)$ : $x = 2$ gives $A = \frac{2}{5};$ $x = - 3$ gives $B = \frac{3}{5} .$
2. $\frac{3}{x^{2} - 4}$
  Show solution
  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
Split into partial fractions and integrate.
1. $\frac{x + 3}{x^{3} + x^{2} - x - 1}$
  Show solution
  Answer $\log ∣ \frac{x - 1}{x + 1} ∣ + \frac{1}{x + 1} + C$
  Factor: $x^{2} (x + 1) - (x + 1) = (x - 1) (x + 1)^{2} .$
  $x + 3 = A (x + 1)^{2} + B (x - 1) (x + 1) + C (x - 1)$ : $x = 1$ gives $A = 1;$ $x = - 1$ gives $C = - 1;$ $x^{2}$ gives $A + B = 0,$ $B = - 1.$
2. $\frac{3 x - 1}{(x^{2} - 1) (x + 2)}$
  Show solution
  Answer $\frac{1}{3} \log ∣ x - 1 ∣ + 2 \log ∣ x + 1 ∣ - \frac{7}{3} \log ∣ x + 2 ∣ + C$
  Three 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
Integrate each function.
1. $\frac{\sin x}{(1 + \cos x) (2 + \cos x)}$
  Show solution
  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. $\frac{(x^{2} + 2) (x^{2} + 5)}{(x^{2} + 1) (x^{2} + 4)}$
  Show solution
  Answer $x + \frac{4}{3} \tan^{- 1} x + \frac{1}{3} \tan^{- 1} \frac{x}{2} + C$
  With $y = x^{2}$ : $\frac{y^{2} + 7 y + 10}{y^{2} + 5 y + 4} = 1 + \frac{2 y + 6}{(y + 1) (y + 4)} .$
  $2 y + 6 = A (y + 4) + B (y + 1)$ : $A = \frac{4}{3},$ $B = \frac{2}{3} .$ So the integrand is $1 + \frac{4 / 3}{x^{2} + 1} + \frac{2 / 3}{x^{2} + 4} .$
Quick answerMediumNCERT Exercise 7.5, Q22
$\int \frac{x d x}{(x + 1) (x + 2)}$ equals
$\log ∣ \frac{(x + 1)^{2}}{x + 2} ∣ + C$ $\log ∣ \frac{(x + 2)^{2}}{x + 1} ∣ + C$ $\log ∣ (x + 1) (x + 2) ∣ + C$ $\log {∣ \frac{x + 2}{x + 1} ∣}^{2} + C$
Show answer
Answer (B) $\log ∣ \frac{(x + 2)^{2}}{x + 1} ∣ + C$
$\frac{x}{(x + 1) (x + 2)} = \frac{- 1}{x + 1} + \frac{2}{x + 2},$ so the integral is $2 \log ∣ x + 2 ∣ - \log ∣ x + 1 ∣ = \log ∣ \frac{(x + 2)^{2}}{x + 1} ∣ .$ Option A swaps the roles of the two factors.
Quick answerMediumNCERT Exercise 7.5, Q23
$\int \frac{d x}{x (x^{2} + 4)}$ equals
$\frac{1}{4} \log ∣ x ∣ + \frac{1}{8} \log (x^{2} + 4) + C$ $\log ∣ x ∣ - \frac{1}{2} \log (x^{2} + 4) + C$ $\frac{1}{4} \log ∣ x ∣ - \frac{1}{8} \log (x^{2} + 4) + C$ $\frac{1}{2} \tan^{- 1} \frac{x}{2} + C$
Show answer
Answer (C) $\frac{1}{4} \log ∣ x ∣ - \frac{1}{8} \log (x^{2} + 4) + C$
$1 = A (x^{2} + 4) + (B x + C) x$ gives $A = \frac{1}{4},$ $B = - \frac{1}{4},$ $C = 0.$ So the integrand is $\frac{1}{4 x} - \frac{x}{4 (x^{2} + 4)},$ which integrates to $\frac{1}{4} \log ∣ x ∣ - \frac{1}{8} \log (x^{2} + 4) .$

7.6 Integration by Parts

Substitution undoes the chain rule. Integration by parts (खंडशः समाकलन) undoes the product rule, and it is the main tool for products such as $x \cos x,$ $x^{2} e^{x}$ or $e^{x} \sin x,$ and for lone functions such as $\log x$ or $\tan^{- 1} x .$

DerivationThe integration by parts formula

Let $u$ and $v$ be differentiable functions of $x .$ The product rule says $\frac{d}{d x} (u v) = u \frac{d v}{d x} + v \frac{d u}{d x} .$ Integrate both sides; the left side gives back $u v$ :

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

Rearranging gives the formula in its shortest form, often written $\int u d v = u v - \int v d u$ :

\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)

To use it on a product $f (x) g (x),$ call $f$ the first function ( $u = f$ ) and $g$ the second function ( $\frac{d v}{d x} = g,$ so $v = \int g d x$ ). Then

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}

In words: keep the first function, multiply by the integral of the second; then subtract the integral of (derivative of the first) × (integral of the second). The method pays off when the new integral is simpler than the old one. Fig. 7.7 shows the formula as a statement about areas.

4/4Together they make $u_{2} v_{2} - u_{1} v_{1}$ : so $\int u d v = [u v] - \int v d u .$

Fig. 7.7Integration by parts as a sum of two areas.

Axes labelled u horizontally and v vertically. In the first step a rising curve runs from the point (u1, v1) to (u2, v2). In the second step the region between the curve and the u-axis, from u1 to u2, is shaded and labelled integral of v du. In the third step the region between the curve and the v-axis, from v1 to v2, is shaded with a different hatch and labelled integral of u dv. In the fourth step both shaded regions are shown together: they fill the big rectangle of area u2 v2 except the small rectangle u1 v1 in the corner, so the two integrals add up to u2 v2 minus u1 v1.

Example 17NCERT Example 17 · p. 260

Choosing the first function

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

Try it yourself first. Then reveal the solution one step at a time.

Choose $u$ so that it gets simpler when differentiated. Logarithms and inverse trigonometric functions become algebraic, and powers of $x$ go down. Exponentials and sines and cosines stay just as complicated, so they make good second functions. A common memory aid is the order inverse trig, log, algebraic, trig, exponential: the earlier type is usually the first function.
No constant is needed in $\int g d x$ . If we use $v = - \frac{1}{2} \cos 2 x + k$ in Example 17, the extra terms are $k x$ from $u v$ and $- \int k d x = - k x$ from the integral, and they cancel.
Parts does not always help. It needs a second function we can integrate and a new integral that is simpler. For $\int \frac{\sin x}{x} d x$ no choice works: this integral has no answer in terms of elementary functions at all.

Example 18NCERT Example 18 · p. 261

One as the second function

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

Try it yourself first. Then reveal the solution one step at a time.

Example 19NCERT Example 19 · p. 261

A power times an exponential

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

Try it yourself first. Then reveal the solution one step at a time.

Example 20NCERT Example 20 · p. 261

The second function needs a substitution

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

Try it yourself first. Then reveal the solution one step at a time.

Example 21NCERT Example 21 · p. 262

When the integral comes back

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

Try it yourself first. Then reveal the solution one step at a time.

7.6.1 Integral of the type ∫eˣ[f(x) + f′(x)] dx

TheoremA product-rule pattern

If $f$ is differentiable, then $\int e^{x} [f (x) + f^{'} (x)] d x = e^{x} f (x) + C .$

The skill is spotting the pattern: inside the bracket, one term must be the derivative of the other.

Example 22NCERT Example 22 · p. 263

Spotting f and f′

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

Try it yourself first. Then reveal the solution one step at a time.

Practice · 5 questions

Practice 7.6: questions with answers

1
Short answerEasyNCERT Exercise 7.6, Q1–Q8
Integrate by parts.
1. $x \cos x$
  Show solution
  Answer $x \sin x + \cos x + C$
  $u = x,$ $v = \sin x$ : $x \sin x - \int \sin x d x .$
2. $x \cos 4 x$
  Show solution
  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
Integrate by parts.
1. $x \cot^{- 1} x$
  Show solution
  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. $(\cos^{- 1} x)^{2}$
  Show solution
  Answer $x (\cos^{- 1} x)^{2} - 2 \sqrt{1 - x^{2}} \cos^{- 1} x - 2 x + C$
  $u = (\cos^{- 1} x)^{2},$ $v = x$ : $x (\cos^{- 1} x)^{2} + 2 \int \frac{x \cos^{- 1} x}{\sqrt{1 - x^{2}}} d x .$
  In the new integral take $\cos^{- 1} x$ first and $\frac{x}{\sqrt{1 - x^{2}}}$ second (its integral is $- \sqrt{1 - x^{2}}$ ): it equals $- \sqrt{1 - x^{2}} \cos^{- 1} x - x .$
3
Short answerMediumNCERT Exercise 7.6, Q16–Q22
Use $\int e^{x} [f (x) + f^{'} (x)] d x = e^{x} f (x) + C,$ or parts twice.
1. $e^{x} (\cos x - \sin x)$
  Show solution
  Answer $e^{x} \cos x + C$
  $f (x) = \cos x,$ $f^{'} (x) = - \sin x .$
2. $\frac{(x + 2) e^{x}}{(x + 3)^{2}}$
  Show solution
  Answer $\frac{e^{x}}{x + 3} + C$
  $\frac{x + 2}{(x + 3)^{2}} = \frac{1}{x + 3} - \frac{1}{(x + 3)^{2}};$ with $f = \frac{1}{x + 3}$ the second term is $f^{'} .$
Quick answerEasyNCERT Exercise 7.6, Q23
$\int x^{3} e^{x^{4}} d x$ equals
$\frac{1}{4} e^{x^{4}} + C$ $\frac{1}{3} e^{x^{4}} + C$ $\frac{1}{4} e^{x^{3}} + C$ $\frac{x^{4}}{4} e^{x^{4}} + C$
Show answer
Answer (A) $\frac{1}{4} e^{x^{4}} + C$
With $t = x^{4},$ $d t = 4 x^{3} d x,$ so the integral is $\frac{1}{4} \int e^{t} d t = \frac{1}{4} e^{x^{4}} + C .$
Quick answerEasyNCERT Exercise 7.6, Q24
$\int e^{x} (\cot x - {c o s e c}^{2} x) d x$ equals
$- e^{x} \cot x + C$ $e^{x} c o s e c x + C$ $e^{x} \tan x + C$ $e^{x} \cot x + C$
Show answer
Answer (D) $e^{x} \cot x + C$
$f = \cot x$ has $f^{'} = - {c o s e c}^{2} x,$ so the bracket is $f + f^{'}$ and the integral is $e^{x} \cot x + C .$

7.6.2 Integrals of some more types

Three square-root integrands complete our list. Here $a > 0.$

Table 7.5Integrals of square roots of quadraticsNCERT 7.6.2 (i)–(iii) · p. 264

	Integral (add $C$ )
(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) has a picture behind it. The graph of $y = \sqrt{a^{2} - x^{2}}$ is the upper half of the circle of radius $a,$ so for $0 \leq x \leq a$ the area under it from $0$ to $x$ splits into a right triangle and a circular sector (Fig. 7.8).

3/3The sector has angle $θ$ with $\sin θ = \frac{x}{a}$ : area $\frac{a^{2}}{2} θ = \frac{a^{2}}{2} \sin^{- 1} \frac{x}{a} .$

Fig. 7.8The area under a circle: a triangle plus a sector.

A quarter circle of radius a in the first quadrant, centre at the origin O. In the first step the region under the arc between x equals 0 and a vertical line at x is shaded, with the top of the vertical line at the point P on the circle. In the second step a triangle with corners O, the foot of the vertical line, and P is highlighted; its base is x and its height is the square root of a squared minus x squared. In the third step the remaining piece, a sector between the vertical axis and the radius OP with angle theta at O, is highlighted; sin theta equals x over a.

Adding the two areas gives exactly the right-hand side of (iii), and it vanishes at $x = 0$ as an area from $0$ to $0$ must. A cleaner proof, valid for all $∣ x ∣ \leq a,$ is to substitute $x = a \sin θ$ : then $\int \sqrt{a^{2} - x^{2}} d x = a^{2} \int \cos^{2} θ d θ = \frac{a^{2}}{2} (θ + \sin θ \cos θ),$ which is the same expression. For (ii) we use parts with $1$ as the second function:

DerivationProof of formula (ii)

Let $I = \int \sqrt{x^{2} + a^{2}} d x .$ Integrating by parts with $\sqrt{x^{2} + a^{2}}$ first and $1$ second,

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

Write $x^{2} = (x^{2} + a^{2}) - a^{2}$ in the last numerator. The integral splits into $I$ itself and $a^{2}$ times formula (6), which we abbreviate as $L = \log ∣ x + \sqrt{x^{2} + a^{2}} ∣$ :

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

So $2 I = x \sqrt{x^{2} + a^{2}} + a^{2} L,$ which gives (ii). Formula (i) follows in exactly the same way using formula (4), and the substitutions $x = a \tan θ,$ $x = a \sec θ$ give the same results.

Example 23NCERT Example 23 · p. 265

Complete the square, then use (ii)

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

Try it yourself first. Then reveal the solution one step at a time.

Example 24NCERT Example 24 · p. 265

Complete the square, then use (iii)

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

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 2

Which formula finishes

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

after completing the square?

(i), with

a = 3

(ii), with

a = 3

(iii), with

a = 3

(iii), with

a = 9

Show answer

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

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

: a constant minus a square, so (iii) with

a = 3

(not

a = 9

a^{2} = 9

Practice · 4 questions

Practice 7.7: questions with answers

1
Short answerMediumNCERT Exercise 7.7, Q1–Q8
Integrate each function with formulas (i)–(iii).
1. $\sqrt{9 - x^{2}}$
  Show solution
  Answer $\frac{x}{2} \sqrt{9 - x^{2}} + \frac{9}{2} \sin^{- 1} \frac{x}{3} + C$
  Formula (iii) with $a = 3.$
2. $\sqrt{4 - 9 x^{2}}$
  Show solution
  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 $\sqrt{4 + \frac{x^{2}}{4}} .$
Show solution
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) with $a = 4$ : $\frac{1}{2} [\frac{x}{2} \sqrt{x^{2} + 16} + 8 \log ∣ x + \sqrt{x^{2} + 16} ∣] + C .$
Quick answerEasyNCERT Exercise 7.7, Q10
$\int \sqrt{x^{2} + 9} d x$ is equal to
$\frac{2}{3} (x^{2} + 9)^{3 / 2} + C$ $\frac{x}{2} \sqrt{x^{2} + 9} + \frac{9}{2} \log ∣ x + \sqrt{x^{2} + 9} ∣ + C$ $\frac{x}{2} \sqrt{x^{2} + 9} + \frac{9}{2} \sin^{- 1} \frac{x}{3} + C$ $\frac{x}{2} \sqrt{x^{2} + 9} + \frac{3}{2} \log ∣ x + \sqrt{x^{2} + 9} ∣ + C$
Show answer
Answer (B) $\frac{x}{2} \sqrt{x^{2} + 9} + \frac{9}{2} \log ∣ x + \sqrt{x^{2} + 9} ∣ + C$
Formula (ii): $\frac{x}{2} \sqrt{x^{2} + a^{2}} + \frac{a^{2}}{2} \log ∣ x + \sqrt{x^{2} + a^{2}} ∣$ with $a^{2} = 9.$ Option A treats the root like a power of $x;$ option D uses $a$ instead of $a^{2} .$
Quick answerMediumNCERT Exercise 7.7, Q11
$\int \sqrt{x^{2} + 6 x + 5} d x$ is equal to
$\frac{x + 3}{2} \sqrt{x^{2} + 6 x + 5} + 2 \log ∣ x + 3 + \sqrt{x^{2} + 6 x + 5} ∣ + C$ $\frac{x + 3}{2} \sqrt{x^{2} + 6 x + 5} - 4 \log ∣ x + 3 + \sqrt{x^{2} + 6 x + 5} ∣ + C$ $\frac{x + 3}{2} \sqrt{x^{2} + 6 x + 5} - 2 \log ∣ x + 3 + \sqrt{x^{2} + 6 x + 5} ∣ + C$ $\frac{x - 3}{2} \sqrt{x^{2} + 6 x + 5} - 2 \log ∣ x - 3 + \sqrt{x^{2} + 6 x + 5} ∣ + C$
Show answer
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) with $y = x + 3,$ $a = 2$ : the log term carries $- \frac{a^{2}}{2} = - 2.$ Option A has the sign of formula (ii); option B forgets to halve $a^{2} .$

7.7 Definite Integral

Now we turn to the second question of Section 7.1, the area under a graph. Suppose $f$ stays above the $x$ -axis on $[a, b] .$ The region between $y = f (x),$ the axis and the vertical lines $x = a$ and $x = b$ has some area; that number, rather than a family of functions, is the definite integral of $f$ from $a$ to $b$ , written $\int_{a}^{b} f (x) d x .$ The numbers $a$ and $b$ are the limits of integration: $a$ at the foot of the integral sign, $b$ at the top.

Definition

Definite integral

For $f$ continuous on $[a, b],$ $\int_{a}^{b} f (x) d x$ is the signed area between the graph of $f$ and the $x$ -axis from $x = a$ to $x = b$ : area above the axis counts positive, area below counts negative. If $F$ is an antiderivative of $f$ on $[a, b],$ its value is $F (b) - F (a) .$

Example $\int_{0}^{2} x d x$ is the area of a triangle with base $2$ and height $2,$ which is $2;$ and $\frac{2^{2}}{2} - \frac{0^{2}}{2} = 2$ as well.

How can such an area be found? One way is to approximate it: cut $[a, b]$ into $n$ thin strips, replace each by a rectangle, and add. As $n$ grows the rectangles hug the curve and the total settles on one value, the limit of a sum (Fig. 7.9). The other way skips the sums: find an antiderivative $F$ and compute $F (b) - F (a) .$ Section 7.8 explains why the two always agree, and from then on we use the second, far quicker, route.

2/3Eight rectangles: the gaps shrink.

Fig. 7.9Adding thinner and thinner rectangles approaches the area under the curve.

A rising curve over the interval from a to b on the x-axis, with the true area under it outlined. In the first step four rectangles of equal width stand under the curve, each as tall as the curve at its left edge, leaving visible gaps under the curve. In the second step there are eight thinner rectangles and the gaps are smaller. In the third step there are thirty-two very thin rectangles and they almost fill the region exactly.

InteractiveArea explorer

An interactive graph with four functions to choose from: x squared, the square root of x, sine x and one over x. Sliders set the interval ends a and b. In Sums mode a slider sets the number of strips from 1 to 64 and buttons choose left, right or midpoint rectangles; the panel shows the rectangle total, the exact value F of b minus F of a from an antiderivative, and the difference between them, which shrinks as the strips get thinner. In Area function mode a slider moves a point x from a to b; the region under the curve from a to x is shaded, a second small graph below plots the area A against x as it is traced, and a short tangent at the current point of that graph has slope equal to f of x, with both numbers shown side by side.

Fig. 7.10Pick a function and an interval. In Sums mode change the number of strips and compare the rectangle total with

F (b) - F (a);

in Area function mode move

x

and watch

A (x)

grow at the rate

f (x) .

Quick check

1 of 3

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

\frac{1}{2},

the area of a triangle

0,

because the interval ends at

0

- \frac{1}{2},

because the region lies below the

x

-axisundefined, because

x

is negative there

Show answer

Answer (C) $- \frac{1}{2},$ because the region lies below the $x$ -axis

[- 1, 0]

the line

y = x

lies below the axis, so the triangle of area

\frac{1}{2}

counts negatively:

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

7.8 Fundamental Theorem of Calculus

7.8.1 Area function

Keep the left end $a$ fixed and slide the right end along the axis. Each position $x$ of the right end encloses its own amount of area under $y = f (t),$ so the area becomes a quantity that varies with $x .$

Definition

Area function

For $f$ continuous on $[a, b],$ the area function is $A (x) = \int_{a}^{x} f (t) d t$ for $a \leq x \leq b .$ It is the area under the graph from $a$ to $x$ (signed, if $f$ changes sign), and $A (a) = 0.$

ExampleFor $f (t) = 2$ and $a = 0$ the region is a rectangle, so $A (x) = 2 x .$

We wrote the variable inside the integral as $t$ so that it is not confused with the moving end $x;$ the letter used inside does not change the value (this is property P0 in Section 7.10). Try the Area function mode of Fig. 7.10: as $x$ moves right, $A (x)$ grows quickly where the curve is high and slowly where it is low.

7.8.2 First fundamental theorem of integral calculus

Theorem 1First fundamental theorem

Let $f$ be continuous on $[a, b]$ and $A (x) = \int_{a}^{x} f (t) d t .$ Then $A^{'} (x) = f (x)$ for every $x$ in $[a, b] .$

3/3The strip is almost a rectangle of height $f (x)$ and width $h,$ so $\frac{A (x + h) - A (x)}{h} \to f (x) .$

Fig. 7.11Why the area function has derivative f(x).

A curve above the x-axis. In the first step the region under it from a to x is shaded and labelled A of x. In the second step x moves right by a small amount h and the new thin strip between x and x plus h is shaded in a different pattern. In the third step a rectangle of width h and height f of x is outlined over the strip, matching it almost exactly; the tiny gap at the top is the only difference.

In words: the rate at which the area grows is the height of the curve. So the area function is an antiderivative of $f .$ This is the bridge between the two problems of Section 7.1.

7.8.3 Second fundamental theorem of integral calculus

Theorem 2Second fundamental theorem

Let $f$ be continuous on $[a, b]$ and let $F$ be any antiderivative of $f$ on $[a, b] .$ Then $\int_{a}^{b} f (x) d x = F (b) - F (a) .$

Evaluating a definite integral

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

Any antiderivative works. With $F + C$ instead of $F$ we get $[F (b) + C] - [F (a) + C],$ and $C$ cancels. So we never write $C$ in a definite integral.
No sums are needed. The value comes from an antiderivative, not from adding rectangles. The hard part is the same as before: finding $F .$
The integrand must be continuous on the whole of $[a, b]$ . Ignoring this gives nonsense. For $\int_{- 1}^{2} \frac{d x}{x^{2}},$ blindly using $F (x) = - \frac{1}{x}$ gives $- \frac{1}{2} - 1 = - \frac{3}{2},$ a negative “area” for a positive integrand. The theorem does not apply, because $\frac{1}{x^{2}}$ is not defined at $x = 0,$ which lies inside $[- 1, 2] .$

Step by step

Working rule for a definite integral

Step 1 of 3

Antiderivative

Find $\int f (x) d x = F (x),$ by any method from this chapter, without the constant.

1 / 3

Example 25NCERT Example 25 · p. 269

Four definite integrals

Evaluate (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) $\int_{0}^{π / 6} \cos^{2} 3 t \sin 3 t d t .$

Try it yourself first. Then reveal the solution one step at a time.

Practice · 5 questions

Practice 7.8: questions with answers

1
Short answerEasyNCERT Exercise 7.8, Q1–Q8
Evaluate each definite integral.
1. $\int_{- 2}^{1} (2 x + 3) d x$
  Show solution
  Answer $6$
  $[x^{2} + 3 x]_{- 2}^{1} = (1 + 3) - (4 - 6) = 6.$
2. $\int_{1}^{e} \frac{2}{x} d x$
  Show solution
  Answer $2$
  $[2 \log x]_{1}^{e} = 2 - 0.$
2
Short answerMediumNCERT Exercise 7.8, Q9–Q15
Evaluate each definite integral.
1. $\int_{0}^{1 / 2} \frac{d x}{\sqrt{1 - x^{2}}}$
  Show solution
  Answer $\frac{π}{6}$
  $[\sin^{- 1} x]_{0}^{1 / 2} .$
2. $\int_{0}^{2} \frac{d x}{4 + x^{2}}$
  Show solution
  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
Evaluate each definite integral.
1. $\int_{1}^{2} \frac{x^{2}}{x^{2} + 3 x + 2} d x$
  Show solution
  Answer $1 + \log \frac{3}{2} - 4 \log \frac{4}{3}$
  Divide and split: $\frac{x^{2}}{(x + 1) (x + 2)} = 1 + \frac{1}{x + 1} - \frac{4}{x + 2} .$
  Antiderivative $x + \log (x + 1) - 4 \log (x + 2);$ from $1$ to $2$ : $1 + \log \frac{3}{2} - 4 \log \frac{4}{3} .$
2. $\int_{0}^{π / 3} (\sec^{2} x + 2 x + 1) d x$
  Show solution
  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
$\int_{1 / \sqrt{3}}^{1} \frac{d x}{1 + x^{2}}$ equals
$\frac{π}{3}$ $\frac{π}{6}$ $\frac{π}{4}$ $\frac{π}{12}$
Show answer
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
$\int_{0}^{1 / 2} \frac{d x}{1 + 4 x^{2}}$ equals
$\frac{π}{8}$ $\frac{π}{4}$ $\frac{π}{2}$ $\frac{π}{16}$
Show answer
Answer (A) $\frac{π}{8}$
$\int \frac{d x}{1 + 4 x^{2}} = \frac{1}{2} \tan^{- 1} 2 x;$ from $0$ to $\frac{1}{2}$ this is $\frac{1}{2} \cdot \frac{π}{4} = \frac{π}{8} .$ Option B forgets the factor $\frac{1}{2} .$

7.9 Evaluation of Definite Integrals by Substitution

To evaluate $\int_{a}^{b} f (x) d x$ with a substitution, we can find the indefinite integral, return to $x,$ and then use the limits $a$ and $b .$ It is usually quicker to change the limits along with the variable and never return to $x,$ as Fig. 7.12 shows.

2/3 $t = x^{3} + 1$ sends $x = 0$ to $t = 1$ and $x = 2$ to $t = 9.$

Fig. 7.12When the variable changes, the limits change with it.

Two horizontal number lines, x on top and t below. In the first step the interval from 0 to 2 on the x line is highlighted. In the second step arrows labelled t equals x cubed plus 1 carry the points 0, 1 and 2 on the x line to the points 1, 2 and 9 on the t line; the arrows spread out because the map stretches. In the third step the interval from 1 to 9 on the t line is highlighted and the integral is rewritten with limits 1 and 9.

Step by step

Working rule for substitution in a definite integral

Step 1 of 3

Substitute

Choose $t = g (x)$ and write $d t = g^{'} (x) d x,$ exactly as for an indefinite integral.

1 / 3

Example 26NCERT Example 26 · p. 272

Two routes, one value

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

Try it yourself first. Then reveal the solution one step at a time.

Example 27NCERT Example 27 · p. 272

An inverse sine as the new variable

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

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

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

put

t = x^{2} - 1.

The new limits are

1

2

1

4

0

1

0

3

Show answer

Answer (D) $0$ to $3$

x = 1

gives

t = 0

and

x = 2

gives

t = 3.

Keeping the old limits

1

and

2

(option A) is the classic slip.

Practice · 3 questions

Practice 7.9: questions with answers

1
Short answerMediumNCERT Exercise 7.9, Q1–Q8
Evaluate by substitution, changing the limits.
1. $\int_{0}^{1} \frac{x^{2}}{x^{3} + 1} d x$
  Show solution
  Answer $\frac{1}{3} \log 2$
  $t = x^{3} + 1$ : limits $1$ to $2;$ $\frac{1}{3} \int_{1}^{2} \frac{d t}{t} .$
2. $\int_{0}^{π / 2} \sqrt{\cos ϕ} \sin^{3} ϕ d ϕ$
  Show solution
  Answer $\frac{8}{21}$
  $\sin^{3} ϕ = (1 - \cos^{2} ϕ) \sin ϕ;$ $t = \cos ϕ,$ limits $1$ to $0$ : $\int_{0}^{1} t^{1 / 2} (1 - t^{2}) d t = \frac{2}{3} - \frac{2}{7} .$
Quick answerHardNCERT Exercise 7.9, Q9
The value of $\int_{1}^{2} \frac{\sqrt{x^{2} - 1}}{x^{4}} d x$ is
$\frac{\sqrt{3}}{4}$ $\frac{3}{8}$ $0$ $\frac{\sqrt{3}}{8}$
Show answer
Answer (D) $\frac{\sqrt{3}}{8}$
$d t = \frac{2}{x^{3}} d x;$ limits $0$ to $\frac{3}{4} .$ So the integral is $\frac{1}{2} \int_{0}^{3 / 4} t^{1 / 2} d t = \frac{1}{3} (\frac{3}{4})^{3 / 2} = \frac{1}{3} \cdot \frac{3 \sqrt{3}}{8} = \frac{\sqrt{3}}{8} .$
Quick answerEasyNCERT Exercise 7.9, Q10
If $f (x) = \int_{0}^{x} t^{2} \cos t d t,$ then $f^{'} (x)$ is
$2 x \cos x - x^{2} \sin x$ $x^{2} \cos x$ $x^{2} \sin x$ $x^{2} \cos x - 1$
Show answer
Answer (B) $x^{2} \cos x$
By Theorem 1 the derivative of $\int_{0}^{x} g (t) d t$ is $g (x),$ here $x^{2} \cos x .$ Option A differentiates the integrand instead.

7.10 Some Properties of Definite Integrals

Some definite integrals are hard or impossible to do through an antiderivative, yet easy once we use the symmetry of the interval or of the integrand. The eight properties in Table 7.6 are the tools. Each follows from the second fundamental theorem or from a substitution.

Table 7.6Properties of definite integrals (here

\int_{a}^{b} f

is short for

\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	$\int_{a}^{b} f (x) d x$ is unchanged if $x$ is replaced by $a + b - x$
P4	P3 for $[0, a]$ : replace $x$ by $a - x$
P5	$\int_{0}^{2 a} f = \int_{0}^{a} [f (x) + f (2 a - x)] d x$
P6	When $f (2 a - x) = f (x)$ (mirror symmetry about $x = a$ ), $\int_{0}^{2 a} f = 2 \int_{0}^{a} f .$ When $f (2 a - x) = - f (x),$ the two halves cancel to $0$
P7	$\int_{- a}^{a} f$ is $2 \int_{0}^{a} f$ for even $f,$ and $0$ for odd $f$

PropertyP0, P1, P2: renaming, reversing, splitting

The name of the variable does not affect the value (P0). Reversing the limits reverses the sign, and an interval of zero width gives $0$ (P1). An integral can be split at any point $c$ (P2).

PropertyP3 and P4: reflecting the interval

Replacing $x$ by $a + b - x$ inside $\int_{a}^{b}$ leaves the value unchanged (P3); for an interval $[0, a]$ the replacement is $x \to a - x$ (P4).

In $\int_{a}^{b} f (x) d x$ put $x = a + b - t,$ so $d x = - d t .$ When $x = a,$ $t = b;$ when $x = b,$ $t = a .$ The integral becomes $- \int_{b}^{a} f (a + b - t) d t,$ which equals $\int_{a}^{b} f (a + b - t) d t$

using P1 to swap the limits; P0 lets us write $x$ again for $t .$ Putting $a = 0$ (and calling the upper limit $a$ ) gives P4. Geometrically, $f (a + b - x)$ is the mirror image of $f (x)$ in the line through the middle of the interval, so it encloses the same area (Fig. 7.13).

PropertyP5 and P6: folding the interval

Over $[0, 2 a]$ the integral equals $\int_{0}^{a} [f (x) + f (2 a - x)] d x$ (P5). So it doubles $\int_{0}^{a} f$ when $f$ is symmetric about $x = a,$ and vanishes when $f$ is antisymmetric about $x = a$ (P6).

PropertyP7: even and odd functions

On an interval $[- a, a]$ that is symmetric about $0,$ an even integrand gives twice the integral over $[0, a]$ and an odd integrand gives $0.$

2/3An odd function's areas are equal and opposite: $\int_{- a}^{a} f = 0.$

Fig. 7.13Symmetry can evaluate an integral without an antiderivative.

In the first step an even curve, symmetric about the y-axis, has equal shaded areas on the left and right of the axis between minus a and a, so the total is twice one side. In the second step an odd curve passes through the origin; the shaded area on the right is above the x-axis and the shaded area on the left is the same shape below it, marked plus and minus, so they cancel. In the third step a curve on an interval from 0 to a is drawn together with its mirror image in the vertical line at a over 2, drawn dashed; both enclose the same area, which is property P4.

Example 28NCERT Example 28 · p. 276

A modulus: split where the sign changes

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

Try it yourself first. Then reveal the solution one step at a time.

3/3Split at $1$ and $2$ : $\frac{5}{6} + \frac{1}{6} + \frac{5}{6} = \frac{11}{6} .$

Fig. 7.14Taking the modulus flips the negative piece upward.NCERT p. 276

A graph of y equals x squared minus 3x plus 2 for x from 0 to 3. In the first step the curve starts at height 2, crosses the x-axis at 1, dips to minus one quarter at 1.5, crosses again at 2 and rises to 2 at x equals 3; the part between 1 and 2 is shaded below the axis and marked minus. In the second step the modulus is taken: the dip is reflected above the axis as a small bump, drawn with a dashed line. In the third step the three pieces are shaded and labelled with their areas five sixths, one sixth and five sixths, which add to eleven sixths.

Example 29NCERT Example 29 · p. 276

An even integrand

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

Try it yourself first. Then reveal the solution one step at a time.

Example 30NCERT Example 30 · p. 277

Getting rid of the x in front

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

Try it yourself first. Then reveal the solution one step at a time.

Example 31NCERT Example 31 · p. 277

An odd integrand

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

Try it yourself first. Then reveal the solution one step at a time.

Example 32NCERT Example 32 · p. 278

Add the mirror image

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

Try it yourself first. Then reveal the solution one step at a time.

Example 33NCERT Example 33 · p. 278

P3 on an interval not starting at 0

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

Try it yourself first. Then reveal the solution one step at a time.

Example 34NCERT Example 34 · p. 279

A famous logarithmic integral

Evaluate $J = \int_{0}^{π} \log \sin x d x,$ and deduce $\int_{0}^{π / 2} \log \sin x d x .$

Try it yourself first. Then reveal the solution one step at a time.

Practice · 6 questions

Practice 7.10: questions with answers

1
Short answerMediumNCERT Exercise 7.10, Q1–Q7
Use symmetry and the properties P0–P7; no long antiderivative is needed.
1. $\int_{0}^{π / 2} \sin^{2} x d x$
  Show solution
  Answer $\frac{π}{4}$
  By P4 it equals $\int_{0}^{π / 2} \cos^{2} x d x .$ Adding, $2 I = \int_{0}^{π / 2} (\sin^{2} x + \cos^{2} x) d x = \frac{π}{2} .$
2. $\int_{0}^{π / 2} \frac{\sqrt[3]{\sin x}}{\sqrt[3]{\sin x} + \sqrt[3]{\cos x}} d x$
  Show solution
  Answer $\frac{π}{4}$
  P4 swaps $\sin$ and $\cos;$ adding the two forms gives $2 I = \frac{π}{2} .$
2
Short answerHardNCERT Exercise 7.10, Q8–Q14
Evaluate each integral.
1. $\int_{0}^{1} \frac{\log (1 + x)}{1 + x^{2}} d x$
  Show solution
  Answer $\frac{π}{8} \log 2$
  Put $x = \tan θ$ : $I = \int_{0}^{π / 4} \log (1 + \tan θ) d θ .$
  By P4, $1 + \tan (\frac{π}{4} - θ) = 1 + \frac{1 - \tan θ}{1 + \tan θ} = \frac{2}{1 + \tan θ},$ so $I = \int_{0}^{π / 4} [\log 2 - \log (1 + \tan θ)] d θ = \frac{π}{4} \log 2 - I .$
2. $\int_{0}^{1} x^{2} \sqrt{1 - x} d x$
  Show solution
  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 or prove.
1. $\int_{0}^{π / 2} \frac{\sin^{2} x - \cos^{2} x}{2 + \sin x \cos x} d x$
  Show solution
  Answer $0$
  P4 swaps $\sin x$ and $\cos x$ : the numerator changes sign, the denominator does not. So $I = - I .$
2. $\int_{0}^{π} \log (1 - \cos x) d x$
  Show solution
  Answer $- π \log 2$
  $1 - \cos x = 2 \sin^{2} \frac{x}{2},$ so $I = π \log 2 + 2 \int_{0}^{π} \log \sin \frac{x}{2} d x .$
  With $x = 2 t$ the last integral is $2 \int_{0}^{π / 2} \log \sin t d t = - π \log 2$ (Example 34). So $I = π \log 2 - 2 π \log 2.$
ProofHardNCERT Exercise 7.10, Q19
If $f (a - x) = f (x)$ and $g (x) + g (a - x) = 6$ for all $x$ in $[0, a],$ show that $\int_{0}^{a} f (x) g (x) d x = 3 \int_{0}^{a} f (x) d x .$
Show solution
Answer $I = 3 \int_{0}^{a} f (x) d x$
By P4, $I = \int_{0}^{a} f (a - x) g (a - x) d x = \int_{0}^{a} f (x) [6 - g (x)] d x = 6 \int_{0}^{a} f (x) d x - I .$
So $2 I = 6 \int_{0}^{a} f (x) d x,$ which is the result.
Quick answerEasyNCERT Exercise 7.10, Q20
The value of $\int_{- 1}^{1} (x^{7} - x \cos x + \sin^{3} x + 2) d x$ is
$0$ $2$ $4$ $8$
Show answer
Answer (C) $4$
$x^{7},$ $x \cos x$ and $\sin^{3} x$ are odd, so they contribute $0.$ The constant $2$ gives $\int_{- 1}^{1} 2 d x = 4.$
Quick answerMediumNCERT Exercise 7.10, Q21
The value of $\int_{0}^{1} \log \frac{1 + 2 x}{3 - 2 x} d x$ is
$\log 3$ $0$ $1$ $2 \log 3$
Show answer
Answer (B) $0$
With $x \to 1 - x,$ $\frac{1 + 2 x}{3 - 2 x}$ becomes $\frac{3 - 2 x}{1 + 2 x},$ so the integrand changes sign: $I = - I,$ hence $I = 0.$

Miscellaneous Examples

Real integrals do not arrive labelled with their method. The examples below mix the techniques of the chapter. Before you calculate, run through the questions in Fig. 7.15: they settle most integrands within a minute.

4/4Is it a product of two different kinds of function? Try integration by parts.

Fig. 7.15Four questions to ask of any integrand, in order.

Four stacked panels joined by downward arrows. In the first step the top panel, standard form, is highlighted with a sample integrand from the table. In the second step the second panel, inner function and its derivative, is highlighted with the sample integrand 2x cos of x squared and the word substitution. In the third step the third panel, ratio of polynomials, is highlighted with a sample rational function and the words partial fractions. In the fourth step the bottom panel, product of two kinds, is highlighted with x times e to the x and the words by parts.

Example 35NCERT Example 35 · p. 281

A substitution under a root

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

Try it yourself first. Then reveal the solution one step at a time.

Example 36NCERT Example 36 · p. 281

Take out the highest power

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

Try it yourself first. Then reveal the solution one step at a time.

Example 37NCERT Example 37 · p. 281

Improper, with a quadratic factor

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

Try it yourself first. Then reveal the solution one step at a time.

Example 38NCERT Example 38 · p. 282

Parts that cancel

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

Try it yourself first. Then reveal the solution one step at a time.

Example 39NCERT Example 39 · p. 283

A double substitution

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

Try it yourself first. Then reveal the solution one step at a time.

Example 40NCERT Example 40 · p. 283

Spot the hidden square

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

Try it yourself first. Then reveal the solution one step at a time.

Example 41NCERT Example 41 · p. 284

A modulus of a product

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

Try it yourself first. Then reveal the solution one step at a time.

Example 42NCERT Example 42 · p. 284

Symmetry, a substitution and a division

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

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

Match each integrand with the first method to try.

Tap an item on the left, then its match on the right.

Show answer

Answer

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

x^{2}

is (a third of) the derivative of

x^{3};

a factorised denominator invites partial fractions; a power times a log calls for parts; a squared sine is reduced by

\sin^{2} θ = \frac{1 - \cos 2 θ}{2} .

Recap

Integrals: key points and summary

Integration reverses differentiation: $\int f (x) d x = F (x) + C$ where $F^{'} = f;$ antiderivatives on an interval differ only by a constant (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).
Every standard derivative gives a standard integral, e.g. $\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C$ ( $n \neq - 1$ ) and $\int \frac{d x}{x} = \log ∣ x ∣ + C$ (Table 7.2).
Substitution: with $t = g (x),$ $\int f (g (x)) g^{'} (x) d x = \int f (t) d t .$ It gives $\int \tan x d x = \log ∣ \sec x ∣ + C$ and $\int \sec x d x = \log ∣ \sec x + \tan x ∣ + C$ (Section 7.3.1).
Identities such as $\cos^{2} x = \frac{1 + \cos 2 x}{2}$ and product-to-sum formulas turn powers and products of sines and cosines into standard forms (Section 7.3.2).
The six special integrals of Table 7.3 handle $\frac{1}{x^{2} \pm a^{2}},$ $\frac{1}{a^{2} - x^{2}}$ and their square roots; completing the square brings any quadratic to them (Section 7.4).
For $\frac{p x + q}{quadratic}$ write $p x + q = A \cdot (derivative of the quadratic) + B$ (Linear numerator).
Partial fractions: divide first if improper; each factor $x - a,$ $(x - a)^{2}$ or irreducible $x^{2} + b x + c$ contributes the terms in Table 7.4 (Section 7.5).
Integration by parts: $\int f g d x = f \int g d x - \int (f^{'} \int g d x) d x;$ choose $f$ so that it simplifies when differentiated (Section 7.6).
$\int e^{x} [f (x) + f^{'} (x)] d x = e^{x} f (x) + C$ (Section 7.6.1); the three square-root integrals are in Table 7.5.
$\int_{a}^{b} f (x) d x$ is the signed area under $y = f (x)$ from $a$ to $b$ (Section 7.7).
First fundamental theorem: $A (x) = \int_{a}^{x} f (t) d t$ has $A^{'} (x) = f (x) .$ Second: $\int_{a}^{b} f (x) d x = F (b) - F (a)$ for any antiderivative $F$ (Section 7.8).
In a definite integral, a substitution changes the limits too (Section 7.9).
Properties P0–P7, especially $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$ and the even/odd rule on $[- a, a],$ often evaluate an integral without an antiderivative (Section 7.10).

Flash cards · 14

Quick recall

1 / 14

Miscellaneous Exercise on Chapter 7

Find your NCERT exercise question

Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
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 on Integrals

7.2.1 Some properties of indefinite integral4 questions
7.3.1 Integration by substitution4 questions
7.3.2 Integration using trigonometric identities3 questions
7.4 Integrals of Some Particular Functions4 questions
7.5 Integration by Partial Fractions3 questions
7.6.1 Integral of the type ∫eˣ[f(x) + f′(x)] dx4 questions
7.6.2 Integrals of some more types2 questions
7.7 Definite Integral3 questions
7.8.3 Second fundamental theorem of integral calculus4 questions
7.9 Evaluation of Definite Integrals by Substitution3 questions
7.10 Some Properties of Definite Integrals4 questions
Miscellaneous Examples3 questions

Glossary · 15

Key terms (English–Hindi)

antiderivativeप्रतिअवकलजprati-avakalaj: A function $F$ whose derivative is $f$ on an interval; also called a primitive of $f$ .First seen in the chapter →
area functionक्षेत्रफल फलनkshetraphal phalan: $A (x) = \int_{a}^{x} f (t) d t$ , the area under the graph from a fixed $a$ up to a moving $x$ .

completing the squareपूर्ण वर्ग बनानाpoorn varg banaana: Rewriting $x^{2} + b x + c$ as ${(x + \frac{b}{2})}^{2} + c - \frac{b^{2}}{4}$ , a square plus a constant.First seen in the chapter →
constant of integrationसमाकलन अचरsamaakalan achar: The arbitrary real number $C$ added to an antiderivative; one extra condition fixes it.First seen in the chapter →

definite integralनिश्चित समाकलनnishchit samaakalan: The number $\int_{a}^{b} f (x) d x$ : the signed area between $y = f (x)$ and the $x$ -axis from $x = a$ to $x = b$ .First seen in the chapter →

even and odd functionsसम और विषम फलनsam aur visham phalan: $f$ is even if $f (- x) = f (x)$ and odd if $f (- x) = - f (x)$ for every $x$ in its domain.First seen in the chapter →

fundamental theorem of calculusकलन की आधारभूत प्रमेयkalan ki aadhaarbhoot prameya: The link between the two halves of calculus: $A^{'} (x) = f (x)$ , and hence $\int_{a}^{b} f (x) d x = F (b) - F (a)$ .First seen in the chapter →

indefinite integralअनिश्चित समाकलनanishchit samaakalan: The family $\int f (x) d x = F (x) + C$ of all antiderivatives of $f$ on an interval.First seen in the chapter →
integrandसमाकल्यsamaakalya: The function $f (x)$ that stands inside $\int f (x) d x$ .First seen in the chapter →
integrationसमाकलनsamaakalan: The process of finding an integral; for indefinite integrals, finding antiderivatives.First seen in the chapter →
integration by partsखंडशः समाकलनkhandashah samaakalan: The rule $\int u v^{'} d x = u v - \int u^{'} v d x$ , the product rule read backwards.First seen in the chapter →
integration by substitutionप्रतिस्थापन द्वारा समाकलनpratisthaapan dvaara samaakalan: Changing the variable, $u = g (x)$ , so that $\int f (g (x)) g^{'} (x) d x$ becomes $\int f (u) d u$ .First seen in the chapter →

limits of integrationसमाकलन की सीमाएँsamaakalan ki seemaaen: The end points $a$ (lower) and $b$ (upper) in $\int_{a}^{b} f (x) d x$ .First seen in the chapter →

partial fractionsआंशिक भिन्नaanshik bhinn: Simpler fractions whose sum equals a given proper rational function.First seen in the chapter →

rational functionपरिमेय फलनparimeya phalan: A quotient $\frac{P (x)}{Q (x)}$ of polynomials; it is proper when $\deg P < \deg Q$ .First seen in the chapter →

Integrals: Class 12 Maths · Chapter 7

What you will learn in Integrals

7.1 Introduction

7.2 Integration as an Inverse Process of Differentiation

7.2.1 Some properties of indefinite integral

Practice 7.1: questions with answers

7.3 Methods of Integration

7.3.1 Integration by substitution

Practice 7.2: questions with answers

7.3.2 Integration using trigonometric identities

Practice 7.3: questions with answers

7.4 Integrals of Some Particular Functions

Practice 7.4: questions with answers

7.5 Integration by Partial Fractions

Practice 7.5: questions with answers

7.6 Integration by Parts

7.6.1 Integral of the type ∫eˣ[f(x) + f′(x)] dx

Practice 7.6: questions with answers

7.6.2 Integrals of some more types

Practice 7.7: questions with answers

7.7 Definite Integral

7.8 Fundamental Theorem of Calculus

7.8.1 Area function

7.8.2 First fundamental theorem of integral calculus

7.8.3 Second fundamental theorem of integral calculus

Practice 7.8: questions with answers

7.9 Evaluation of Definite Integrals by Substitution

Practice 7.9: questions with answers

7.10 Some Properties of Definite Integrals

Practice 7.10: questions with answers

Miscellaneous Examples

Integrals: key points and summary

Miscellaneous Exercise on Chapter 7

Practice questions with answers (NCERT “Miscellaneous exercises”)

Find your NCERT exercise question

Quick check: MCQs on Integrals

Key terms (English–Hindi)

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