Chapter 7 · Integrals · Go for it

Integrals, the quick way

Integration is differentiation run backwards. Learn twelve moves and you can answer most questions in this chapter.

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The chapter in one minute

What integration is, and what the exam asks.

Integration is differentiation run backwards. You are given a derivative, and you find the function it came from.

Many functions have the same derivative (they differ only by a constant), so every answer ends with $+\,C$ ⁠.

A definite integral has limits, like $\int_0^1$ ⁠, and its answer is a number.

Most exam questions on this chapter are one of twelve kinds, and each kind has a fixed recipe. You will learn all twelve below.

12question types

16formula cards

1check that catches most mistakes: differentiate your answer

+ Con every answer without limits

Formulas

The formulas you must know

Learn these cards. Almost everything in the chapter is built from them.

Powers of x

\int x^n\,dx = \frac{x^{n + 1}}{n + 1} + C,\ n \ne -1

Add 1 to the power, then divide by the new power. For the power −1, use the next card.

One over x

\int\frac{1}{x}\,dx = \log|x| + C

This is the power −1. Keep the modulus bars.

Exponential

\int e^x\,dx = e^x + C

$e^x$ is its own integral.

Sine

\int\sin x\,dx = -\cos x + C

Watch the minus sign.

Cosine

\int\cos x\,dx = \sin x + C

Also $\int\sec^2 x\,dx = \tan x + C$ and $\int\sec x\tan x\,dx = \sec x + C$ ⁠.

Sum of two squares

\begin{aligned} &\int\frac{dx}{x^2 + a^2} \\ &= \frac{1}{a}\tan^{-1}\frac{x}{a} + C \end{aligned}

The most used special form. Here $a \ne 0$ ⁠.

Difference of two squares

\begin{aligned} &\int\frac{dx}{x^2 - a^2} \\ &= \frac{1}{2a}\log\left|\frac{x - a}{x + a}\right| + C \end{aligned}

Minus in the bottom gives a log. The other way round, $\int\frac{dx}{a^2 - x^2} = \frac{1}{2a}\log\left|\frac{a + x}{a - x}\right| + C$ ⁠.

Root of

a^2 - x^2

\begin{aligned} &\int\frac{dx}{\sqrt{a^2 - x^2}} \\ &= \sin^{-1}\frac{x}{a} + C \end{aligned}

Square root, number first, then minus. Here $a > 0$ ⁠.

Root of

x^2 \pm a^2

\begin{aligned} &\int\frac{dx}{\sqrt{x^2 \pm a^2}} \\ &= \log\left|x + \sqrt{x^2 \pm a^2}\right| \\ &\quad + C \end{aligned}

Same sign on both sides.

Root on top:

a^2 - x^2

\begin{aligned} &\int\sqrt{a^2 - x^2}\,dx \\ &= \frac{x}{2}\sqrt{a^2 - x^2} \\ &\quad + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C \end{aligned}

Now the root is on top. Here $a > 0$ ⁠.

Root on top:

x^2 \pm a^2

\begin{aligned} &\int\sqrt{x^2 \pm a^2}\,dx \\ &= \frac{x}{2}\sqrt{x^2 \pm a^2} \\ &\quad \pm \frac{a^2}{2}\log\left|x + \sqrt{x^2 \pm a^2}\right| \\ &\quad + C \end{aligned}

The same sign in all three places.

By parts

\begin{aligned} &\int u\,v\,dx \\ &= u\int v\,dx \\ &\quad - \int\left(u'\int v\,dx\right)dx \end{aligned}

To pick $u$ ⁠, use the memory aid ILATE (see below).

The

e^x

pair

\begin{aligned} &\int e^x\big[f(x) + f'(x)\big]dx \\ &= e^x f(x) + C \end{aligned}

Once you spot $f$ and $f'$ ⁠, write them down and the answer follows.

Definite integral

\int_a^b f(x)\,dx = F(b) - F(a)

$F$ is any antiderivative of $f$ ⁠. No $+\,C$ here.

The

f(a - x)

property

\begin{aligned} &\int_0^a f(x)\,dx \\ &= \int_0^a f(a - x)\,dx \end{aligned}

Then add the two forms of the integral.

Odd function:

f(-x) = -f(x)

\int_{-a}^{a} f(x)\,dx = 0

If instead $f(-x) = f(x)$ (even), $\int_{-a}^{a} f(x)\,dx = 2\int_0^a f(x)\,dx$ ⁠.

Question types

The question types and how to crack each one

For each type: how to spot it, what to do, and one real NCERT question worked the exam way.

1. Basic integrals: tidy up, then use the cards

Spot it: powers of $x$ ⁠, $e^x$ ⁠, sin or cos, or a fraction with a single term at the bottom.

Split or expand until every term is on a formula card. With one term at the bottom, divide each term on top by it.
Integrate term by term.
Add one $+\,C$ at the end.

Exercise 7.1 Q11. Find $I = \int\frac{x^3 + 5x^2 - 4}{x^2}\,dx$ ⁠.

\begin{aligned} I &= \int\left(x + 5 - \frac{4}{x^2}\right)dx \\ &= \frac{x^2}{2} + 5x \\ &\quad - 4\cdot\frac{x^{-1}}{-1} + C \end{aligned}

∴ $I = \frac{x^2}{2} + 5x + \frac{4}{x} + C$

See a full exam answer · Go deeper: see it visually · Basics: powers of x

In past papers (1): click to see

Sample 2022-23 Q5

2. Substitution: when the derivative is hiding in the question

Spot it: a function and its derivative both appear, like $\log x$ and $\frac{1}{x}$ ⁠, or $x^2 + 1$ and $x$ ⁠.

Put $t$ equal to the inside function, the one whose derivative you can see.
Write the $dt$ line: $dt$ is its derivative times $dx$ ⁠.
Rewrite everything in $t$ and integrate with a card.
Put $x$ back and add $+\,C$ ⁠.

Exercise 7.2 Q2. Find $I = \int\frac{(\log x)^2}{x}\,dx$ ⁠.

Put $\log x = t$ ⁠. Then $\frac{1}{x}\,dx = dt$ ⁠.

\begin{aligned} I &= \int t^2\,dt \\ &= \frac{t^3}{3} + C \end{aligned}

∴ $I = \frac{(\log x)^3}{3} + C$

See a full exam answer · Go deeper: see it visually · Basics: the chain rule backwards

In past papers (21): click to see

3. Squares and products of sin and cos: use an identity first

Spot it: $\sin^2$ ⁠, $\cos^2$ ⁠, $\sin^3$ ⁠, or a product such as $\sin 3x\cos 4x$ ⁠.

Lower a square: $\sin^2 x = \frac{1 - \cos 2x}{2}$ and $\cos^2 x = \frac{1 + \cos 2x}{2}$ ⁠.
Lower a cube: $\sin^3 x = \frac{3\sin x - \sin 3x}{4}$ ⁠.
Turn a product into a sum: $2\sin A\cos B$ $= \sin(A + B) + \sin(A - B)$ ⁠.
Integrate each term, dividing by the number in front of $x$ ⁠.

Exercise 7.3 Q1. Find $I = \int\sin^2(2x + 5)\,dx$ ⁠.

Using $\sin^2\theta = \frac{1 - \cos 2\theta}{2}$ ⁠,

\begin{aligned} I &= \frac{1}{2}\int\big(1 - \cos(4x + 10)\big)dx \\ &= \frac{1}{2}\left(x - \frac{\sin(4x + 10)}{4}\right) \\ &\quad + C \end{aligned}

∴ $I = \frac{x}{2} - \frac{\sin(4x + 10)}{8} + C$

See a full exam answer · Go deeper: see it visually · Basics: trig identities

4. Special forms: complete the square, then match a card

Spot it: $dx$ over a quadratic, or over the square root of a quadratic, with just a number on top.

Complete the square, so the bottom looks like $(x + p)^2 + k^2$ ⁠, $(x + p)^2 - k^2$ or $k^2 - (x + p)^2$ ⁠.
Put $t$ equal to the bracket, and write the $dt$ line.
Match one of the four special-form cards and read off $a$ ⁠.

Exercise 7.4 Q11. Find $I = \int\frac{dx}{9x^2 + 6x + 5}$ ⁠.

Complete the square: $9x^2 + 6x + 5 = (3x + 1)^2 + 2^2$ ⁠. Put $3x + 1 = t$ ⁠. Then $3\,dx = dt$ ⁠.

\begin{aligned} I &= \frac{1}{3}\int\frac{dt}{t^2 + 2^2} \\ &= \frac{1}{3}\cdot\frac{1}{2}\tan^{-1}\frac{t}{2} + C \end{aligned}

∴ $I = \frac{1}{6}\tan^{-1}\frac{3x + 1}{2} + C$

See a full exam answer · Go deeper: see it visually · Basics: completing the square

In past papers (2): click to see

Board 2024 Q33(a) · Sample 2022-23 Q26

5. A linear top over a quadratic: split the top

Spot it: $px + q$ on top, and a quadratic, or the square root of one, at the bottom.

Write the top as $A$ times the derivative of the quadratic, plus $B$ ⁠. Compare coefficients to find $A$ and $B$ ⁠.
Split into two integrals.
First part: put $t$ equal to the quadratic. It gives a log, or $2\sqrt t$ when the quadratic is under a root.
Second part has just a number on top: finish it as in type 4.

Exercise 7.4 Q17. Find $I = \int\frac{x + 2}{\sqrt{x^2 - 1}}\,dx$ ⁠.

The derivative of $x^2 - 1$ is $2x$ ⁠, so write $x + 2 = \frac{1}{2}(2x) + 2$ ⁠.

\begin{aligned} I &= \frac{1}{2}\int\frac{2x\,dx}{\sqrt{x^2 - 1}} \\ &\quad + 2\int\frac{dx}{\sqrt{x^2 - 1}} \end{aligned}

In the first part, put $x^2 - 1 = t$ ⁠. Then $2x\,dx = dt$ ⁠, and it gives $\sqrt t$ ⁠. The second part uses the root of $x^2 \pm a^2$ card, with the minus sign and 1 in place of $a$ ⁠.

∴ $I = \sqrt{x^2 - 1}$ $+ 2\log\left|x + \sqrt{x^2 - 1}\right| + C$

See a full exam answer · Go deeper: see it visually

In past papers (2): click to see

Board 2026 Q27(a) · Sample 2021-22 T2 Q7

6. A square root on top: use the root-on-top cards

Spot it: you integrate the square root of a quadratic itself, like $\sqrt{4 - x^2}$ or $\sqrt{x^2 + 4x + 6}$ ⁠.

If there is an $x$ term inside the root, complete the square.
Put $t$ equal to the bracket, and write the $dt$ line.
Use the matching root-on-top card and read off $a$ ⁠.
Put $x$ back and add $+\,C$ ⁠.

Exercise 7.7 Q1. Find $I = \int\sqrt{4 - x^2}\,dx$ ⁠.

Here $4 - x^2 = 2^2 - x^2$ ⁠, so use the root-on-top card for $a^2 - x^2$ ⁠, with 2 in place of $a$ ⁠.

\begin{aligned} I &= \frac{x}{2}\sqrt{4 - x^2} \\ &\quad + \frac{4}{2}\sin^{-1}\frac{x}{2} + C \end{aligned}

∴ $I = \frac{x}{2}\sqrt{4 - x^2} + 2\sin^{-1}\frac{x}{2} + C$

See a full exam answer · Go deeper: see it visually · Basics: completing the square

7. Partial fractions: break a fraction into simple pieces

Spot it: a fraction of two polynomials whose bottom splits into factors.

If the top's degree is not smaller than the bottom's, divide first.
Write the form: $\frac{A}{x - p} + \frac{B}{x - q}$ for two different factors, $\frac{A}{x - p} + \frac{B}{(x - p)^2}$ for a repeated factor, $\frac{Ax + B}{x^2 + 1}$ for a factor like $x^2 + 1$ ⁠.
Find $A$ and $B$ ⁠: put $x$ equal to each root, or compare coefficients.
Integrate each piece; most of them give logs.

Exercise 7.5 Q1. Find $I = \int\frac{x}{(x + 1)(x + 2)}\,dx$ ⁠.

Let $\frac{x}{(x + 1)(x + 2)} = \frac{A}{x + 1} + \frac{B}{x + 2}$ ⁠. So $x = A(x + 2) + B(x + 1)$ ⁠.

Put $x = -1$ ⁠: $A = -1$ ⁠. Put $x = -2$ ⁠: $B = 2$ ⁠.

\begin{aligned} I &= -\int\frac{dx}{x + 1} \\ &\quad + 2\int\frac{dx}{x + 2} \\ &= -\log|x + 1| \\ &\quad + 2\log|x + 2| + C \end{aligned}

∴ $I = \log\frac{(x + 2)^2}{|x + 1|} + C$

See a full exam answer · Go deeper: see it visually · Basics: splitting fractions

In past papers (9): click to see

Board 2026 Q27(b) · Board 2024 Q28(a) · Board 2023 Q27 · Board 2022 T2 Q9(b) · Sample 2025-26 Q33(b) · Sample 2023-24 Q26 · Sample 2022-23 Q31 · Sample 2021-22 T2 Q7 · Sample 2020-21 Q33

8. Integration by parts: a product of two different kinds

Spot it: a product such as $x\sin x$ ⁠, $xe^x$ or $x\log x$ ⁠. A lone $\log x$ or $\tan^{-1}x$ works too: take the other function as 1.

To pick $u$ ⁠, use the memory aid ILATE: Inverse trig, Log, Algebraic, Trig, Exponential. The one that comes first is usually $u$ ⁠. It is a guide, not a rule.
Use the by-parts card: $u$ times $\int v$ ⁠, minus the integral of $u'$ times $\int v$ ⁠.
The new integral should be easier. If it is harder, swap your choice.
Finish with $+\,C$ ⁠.

Exercise 7.6 Q1. Find $I = \int x\sin x\,dx$ ⁠.

ILATE puts $x$ (A) before $\sin x$ (T), so take $u = x$ and $v = \sin x$ ⁠.

\begin{aligned} I &= x(-\cos x) \\ &\quad - \int 1\cdot(-\cos x)\,dx \\ &= -x\cos x + \int\cos x\,dx \end{aligned}

∴ $I = -x\cos x + \sin x + C$

See a full exam answer · Go deeper: see it visually

In past papers (12): click to see

Board 2026 Q26 · Board 2025 Q28(a) · Board 2024 Q33(b) · Board 2023 Q5 · Board 2023 Q28(a) · Board 2023 Q31(a) · Sample 2026-27 Q29(a) · Sample 2025-26 Q34(a) · Sample 2024-25 Q14 · Sample 2024-25 Q29(a) · Sample 2023-24 Q29(b) · Sample 2021-22 T2 Q1(a)

9. The $e^x$ shortcut: a function and its derivative in a bracket

Spot it: $e^x$ times a bracket of two parts, where one part is the derivative of the other.

Name $f$ ⁠: the part whose derivative is the other part.
Check that $f'$ really is the other part.
Write the answer straight away: $e^x f(x) + C$ ⁠.

Exercise 7.6 Q19. Find $I = \int e^x\left(\frac{1}{x} - \frac{1}{x^2}\right)dx$ ⁠.

Take $f(x) = \frac{1}{x}$ ⁠. Then $f'(x) = -\frac{1}{x^2}$ ⁠.

\begin{aligned} I &= \int e^x\big[f(x) + f'(x)\big]dx \\ &= e^x f(x) + C \end{aligned}

∴ $I = \frac{e^x}{x} + C$

See a full exam answer · Go deeper: see it visually

In past papers (3): click to see

Board 2022 T2 Q3(b) · Sample 2025-26 Q23(a) · Sample 2020-21 Q7(a)

10. Definite integrals: integrate, then top minus bottom

Spot it: numbers on the integral sign. The answer is a number, with no $+\,C$ ⁠.

Find an antiderivative $F$ ⁠, by any of the moves above.
Work out $F(\text{top}) - F(\text{bottom})$ ⁠.
If you substitute, change the limits too, on their own line, and stay with $t$ to the end.

Exercise 7.9 Q1. Evaluate $I = \int_0^1\frac{x}{x^2 + 1}\,dx$ ⁠.

Put $x^2 + 1 = t$ ⁠. Then $2x\,dx = dt$ ⁠.

When $x = 0$ ⁠, $t = 1$ ⁠; when $x = 1$ ⁠, $t = 2$ ⁠.

\begin{aligned} I &= \frac{1}{2}\int_1^2\frac{dt}{t} = \frac{1}{2}\Big[\log t\Big]_1^2 \\ &= \frac{1}{2}(\log 2 - \log 1) \end{aligned}

∴ $I = \frac{1}{2}\log 2$

See a full exam answer · Go deeper: see it visually

In past papers (3): click to see

Board 2023 Q6 · Sample 2026-27 Q19 · Sample 2022-23 Q9

11. Limits from $-a$ to $a$ ⁠: check odd or even

Spot it: the two limits are the same number with opposite signs.

Work out $f(-x)$ and write that line.
If $f(-x) = -f(x)$ ⁠, the function is odd and the answer is 0.
If $f(-x) = f(x)$ ⁠, it is even: double the integral from 0 to $a$ ⁠.

Exercise 7.10 Q13. Evaluate $I = \int_{-\pi/2}^{\pi/2}\sin^7 x\,dx$ ⁠.

Let $f(x) = \sin^7 x$ ⁠. Then

\begin{aligned} f(-x) &= (\sin(-x))^7 \\ &= (-\sin x)^7 = -f(x) \end{aligned}

So $f$ is odd, and $\int_{-a}^{a} f(x)\,dx = 0$ for an odd function.

∴ $I = 0$

See a full exam answer · Go deeper: see it visually · Basics: symmetry

In past papers (8): click to see

Board 2026 Q9 · Board 2026 Q28 · Board 2025 Q3 · Sample 2026-27 Q19 · Sample 2026-27 Q33 · Sample 2023-24 Q9 · Sample 2023-24 Q24 · Sample 2020-21 Q7(b)

12. Limits from 0 to $a$ ⁠: the $f(a - x)$ trick

Spot it: replacing $x$ by $a - x$ swaps the parts of the question, like sin and cos when $a$ is $\frac{\pi}{2}$ ⁠.

Call the integral $I$ ⁠, and label it (i).
Replace $x$ by $a - x$ everywhere, using the $f(a - x)$ card. Label it (ii).
Add (i) and (ii). The hard part cancels and $2I$ is easy.
Divide by 2.

Exercise 7.10 Q2. Evaluate

I = \int_0^{\pi/2}\frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}}\,dx

This is (i).

Using $\int_0^a f(x)\,dx = \int_0^a f(a - x)\,dx$ ⁠, with $\sin\left(\frac{\pi}{2} - x\right) = \cos x$ and $\cos\left(\frac{\pi}{2} - x\right) = \sin x$ ⁠,

I = \int_0^{\pi/2}\frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}}\,dx

This is (ii). Adding (i) and (ii),

2I = \int_0^{\pi/2}1\,dx = \frac{\pi}{2}

∴ $I = \frac{\pi}{4}$

See a full exam answer · Go deeper: see it visually · Basics: symmetry

In past papers (9): click to see

Board 2023 Q26 · Board 2022 T2 Q13 · Sample 2025-26 Q11 · Sample 2025-26 Q33(a) · Sample 2024-25 Q13 · Sample 2024-25 Q29(b) · Sample 2023-24 Q28(b) · Sample 2022-23 Q28(a) · Sample 2020-21 Q23(b)

Shortcuts

Shortcuts and quick tips

Small habits that save time and marks.

Check by differentiating your answer. If you get the question back, you are right. It takes about 20 seconds.
Tidy up before you integrate: split a fraction, expand a bracket, or use an identity.
Is the inside $ax + b$ ⁠? Integrate as usual and divide by $a$ ⁠: $\int e^{2x + 3}\,dx = \frac{e^{2x + 3}}{2} + C$ ⁠.
Is the top the derivative of the bottom? The answer is the log of the bottom, with modulus bars: $\int\frac{2x}{1 + x^2}\,dx = \log(1 + x^2) + C$ ⁠. Here no bars are needed, because $1 + x^2$ is always positive.
Squares of sin or cos: lower the power first, for example $\cos^2 x = \frac{1 + \cos 2x}{2}$ ⁠.
Definite integrals never need $+\,C$ ⁠: it cancels when you subtract.
Limits from $-a$ to $a$ ⁠? Test for odd first. If $f(-x) = -f(x)$ ⁠, the answer is 0 at once.
In an MCQ, differentiate the options. The one that gives back the question is the answer.

Watch out

Mistakes that cost marks

Each one is common, and each has an easy fix.

Leaving out $+\,C$ → write it once, at the end of every answer without limits.
Writing $\int\cos 3x\,dx = 3\sin 3x$ → divide by 3 instead of multiplying: $\frac{1}{3}\sin 3x + C$ ⁠.
Changing to $t$ but keeping the old $x$ limits → work out the new limits on their own line and use them.
Dropping the modulus bars → keep them: $\int\frac{dx}{x} = \log|x| + C$ ⁠.
Only one fraction for a repeated factor → for $(x - 1)^2$ write $\frac{A}{x - 1} + \frac{B}{(x - 1)^2}$ ⁠.
Partial fractions when the top's degree is not smaller → divide first, then split.
Integrating a product as $\int u\,dx$ times $\int v\,dx$ → that is never the rule; use by parts.
Saying "odd, so 0" with no working → write the $f(-x)$ line; it carries a mark.

Marks

How the examiner gives marks

From the marking schemes CBSE publishes with its sample papers. The mark splits are examples from those papers, not fixed rules. Writing these lines is what earns marks.

The substitution line, with both halves (Put $t$ = …, $dt$ = … $dx$ ⁠), is worth half a mark on its own.
In a definite integral, the new limits on their own line ("when $x$ is 0, $t$ is 1") are another half mark.
Partial fractions: the form with $A$ and $B$ ⁠, their values, the split integrals and the answer each earn marks.
By parts: each line of by-parts working earns marks. In one 3-mark sample question, each of the three lines earned 1 mark.
Properties: write the property next to the line that uses it, and for odd or even show the $f(-x)$ line. Each is about 1 mark.
Give the final answer on its own line, in $x$ ⁠, with $+\,C$ ⁠. A slip is penalised only once, so keep going. An MCQ earns 1 mark for the right option only.

Practice

Practise these first

Easiest first. Together they cover every type above.

Exercise 7.1 Q18 basic cards: sec squared and sec tan
Exercise 7.2 Q1 substitution: the top is the derivative of the bottom
Exercise 7.3 Q2 a product of sin and cos turned into a sum
Exercise 7.6 Q4 by parts with a log (L comes first)
Exercise 7.6 Q16 the $e^x$ shortcut
Exercise 7.10 Q11 an even function between $-a$ and $a$
Exercise 7.9 Q5 a definite integral: substitution with new limits
Exercise 7.4 Q10 complete the square, then a root card
Exercise 7.4 Q22 a linear top: split it into a log and a special form
Exercise 7.7 Q3 complete the square under a root on top
Exercise 7.5 Q8 partial fractions with a repeated factor
Exercise 7.10 Q3 the $f(a - x)$ trick, then add

You can do this

Twelve moves, one at a time. Work through the practice list, check each answer by differentiating, and you are ready for the exam.

Ready to go deeper? Open the interactive lesson

Past papers

46 CBSE board and sample-paper questions on Integrals, each with a full-marks exam answer. Practise them →

Integrals, the quick way

The chapter in one minute

The formulas you must know

The question types and how to crack each one

1. Basic integrals: tidy up, then use the cards

2. Substitution: when the derivative is hiding in the question

3. Squares and products of sin and cos: use an identity first

4. Special forms: complete the square, then match a card

5. A linear top over a quadratic: split the top

6. A square root on top: use the root-on-top cards

7. Partial fractions: break a fraction into simple pieces

8. Integration by parts: a product of two different kinds

9. The exe^xex shortcut: a function and its derivative in a bracket

10. Definite integrals: integrate, then top minus bottom

11. Limits from −a-a−a to aaa⁠: check odd or even

12. Limits from 0 to aaa⁠: the f(a−x)f(a - x)f(a−x) trick

Shortcuts and quick tips

Mistakes that cost marks

How the examiner gives marks

Practise these first

You can do this

9. The $e^x$ shortcut: a function and its derivative in a bracket

11. Limits from $-a$ to $a$ ⁠: check odd or even

12. Limits from 0 to $a$ ⁠: the $f(a - x)$ trick