What is Integration by Substitution in Class 12 Maths?

Integration by Substitution is part of Chapter 7 (Integrals) in NCERT Class 12 Mathematics. This topic is fundamental for CBSE board exams and competitive tests.

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Integration by Substitution

🎓 Class 12 Mathematics CBSE Theory Ch 7 — Integrals ⏱ ~25 min

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7.3 Methods of Integration

In previous sections, we discussed integrals of those functions which were readily obtainable from derivatives of some functions. It was based on inspection, i.e., on the search for a function F whose derivative is \(f\) which led us to the integral of \(f\). However, this method does not work for a large class of functions. Hence, we need to develop additional techniques or methods for finding integrals. Prominent among these are methods based on:

Integration by Substitution^?
Integration using Partial Fractions
Integration by Parts

7.3.1 Integration by Substitution

Key Method

The given integral \(\int f(x)\,dx\) can be transformed into another form by changing the independent variable \(x\) to \(t\) by substituting \(x = g(t)\).

Consider \(I = \int f(x)\,dx\). Put \(x = g(t)\) so that \(\frac{dx}{dt} = g'(t)\).
We write \(dx = g'(t)\,dt\).
Thus \(I = \int f(x)\,dx = \int f(g(t))\cdot g'(t)\,dt\)

This change of variable formula is one of the most important tools available to us in the name of integration by substitution. Usually, we make a substitution for a function whose derivative also occurs in the integrand, as illustrated in the following examples.

Example 5

Integrate the following functions w.r.t. \(x\):

(i) \(\sin mx\) (ii) \(2x\sin(x^2 + 1)\) (iii) \(\frac{\tan^4\sqrt{x}\sec^2\sqrt{x}}{\sqrt{x}}\) (iv) \(\frac{\sin(\tan^{-1}x)}{1 + x^2}\)

Solution

(i) We know that derivative of \(mx\) is \(m\). Thus, we make the substitution \(mx = t\) so that \(m\,dx = dt\). \[\int \sin mx\,dx = \frac{1}{m}\int \sin t\,dt = \frac{1}{m}(-\cos t) + C = -\frac{1}{m}\cos mx + C\]

(ii) Derivative of \(x^2 + 1\) is \(2x\). Thus, we use the substitution \(x^2 + 1 = t\) so that \(2x\,dx = dt\). \[\int 2x\sin(x^2 + 1)\,dx = \int \sin t\,dt = -\cos t + C = -\cos(x^2 + 1) + C\]

(iii) Derivative of \(\sqrt{x}\) is \(\frac{1}{2\sqrt{x}}\). Thus we use the substitution \(\tan\sqrt{x} = t\) so that \(\frac{\sec^2\sqrt{x}}{2\sqrt{x}}\,dx = dt\). \[\int \frac{\tan^4\sqrt{x}\sec^2\sqrt{x}}{\sqrt{x}}\,dx = 2\int t^4\,dt = \frac{2t^5}{5} + C = \frac{2}{5}\tan^5\sqrt{x} + C\]

(iv) Derivative of \(\tan^{-1}x\) is \(\frac{1}{1+x^2}\). Thus, we use the substitution \(\tan^{-1}x = t\) so that \(\frac{dx}{1+x^2} = dt\). \[\int \frac{\sin(\tan^{-1}x)}{1 + x^2}\,dx = \int \sin t\,dt = -\cos t + C = -\cos(\tan^{-1}x) + C\]

Important Integrals Involving Trigonometric Functions

Now we discuss some important integrals involving trigonometric functions and their standard results using substitution technique. These will be used later without reference.

Standard Results

(i) \(\int \tan x\,dx = \log|\sec x| + C\)

(ii) \(\int \cot x\,dx = \log|\sin x| + C\)

(iii) \(\int \sec x\,dx = \log|\sec x + \tan x| + C\)

(iv) \(\int \text{cosec}\,x\,dx = \log|\text{cosec}\,x - \cot x| + C\)

Proof of (i): \(\int \tan x\,dx\)

Derivation

\[\int \tan x\,dx = \int \frac{\sin x}{\cos x}\,dx\] Put \(\cos x = t\) so that \(-\sin x\,dx = dt\), i.e., \(\sin x\,dx = -dt\). \[\int \frac{\sin x}{\cos x}\,dx = -\int \frac{dt}{t} = -\log|t| + C = -\log|\cos x| + C = \log|\sec x| + C\]

Proof of (ii): \(\int \cot x\,dx\)

Derivation

\[\int \cot x\,dx = \int \frac{\cos x}{\sin x}\,dx\] Put \(\sin x = t\) so that \(\cos x\,dx = dt\). \[\int \frac{dt}{t} = \log|t| + C = \log|\sin x| + C\]

Proof of (iii): \(\int \sec x\,dx\)

Derivation

\[\int \sec x\,dx = \int \frac{\sec x(\sec x + \tan x)}{\sec x + \tan x}\,dx\] Put \(\sec x + \tan x = t\) so that \((\sec x\tan x + \sec^2 x)\,dx = dt\). \[\int \frac{dt}{t} = \log|t| + C = \log|\sec x + \tan x| + C\]

Proof of (iv): \(\int \text{cosec}\,x\,dx\)

Derivation

\[\int \text{cosec}\,x\,dx = \int \frac{\text{cosec}\,x(\text{cosec}\,x + \cot x)}{\text{cosec}\,x + \cot x}\,dx\] Put \(\text{cosec}\,x + \cot x = t\) so that \((-\text{cosec}\,x\cot x - \text{cosec}^2 x)\,dx = dt\). Therefore, \(\text{cosec}\,x(\text{cosec}\,x + \cot x)\,dx = -dt\). \[\int \frac{-dt}{t} = -\log|t| + C = -\log|\text{cosec}\,x + \cot x| + C = \log|\text{cosec}\,x - \cot x| + C\]

Example 6

Find the following integrals:

(i) \(\int \sin^3 x\cos^2 x\,dx\) (ii) \(\frac{\sin x}{\sin(x+a)}\) (iii) \(\frac{1}{1 + \tan x}\)

Solution

(i) Put \(t = \cos x\) so that \(dt = -\sin x\,dx\). \[\int \sin^3 x\cos^2 x\,dx = \int \sin^2 x\cdot\cos^2 x\cdot\sin x\,dx = \int (1-\cos^2 x)\cos^2 x\sin x\,dx\] \[= -\int (1-t^2)t^2\,dt = -\int(t^2 - t^4)\,dt = -\frac{t^3}{3} + \frac{t^5}{5} + C = -\frac{\cos^3 x}{3} + \frac{\cos^5 x}{5} + C\]

(ii) Put \(x + a = t\), so \(dx = dt\) and \(x = t - a\). \[\int \frac{\sin x}{\sin(x+a)}\,dx = \int \frac{\sin(t-a)}{\sin t}\,dt = \int \frac{\sin t\cos a - \cos t\sin a}{\sin t}\,dt\] \[= \cos a\int dt - \sin a\int\cot t\,dt = t\cos a - \sin a\cdot\log|\sin t| + C\] \[= (x+a)\cos a - \sin a\cdot\log|\sin(x+a)| + C\]

(iii) \[\int \frac{1}{1+\tan x}\,dx = \int \frac{\cos x}{\cos x + \sin x}\,dx\] Now, consider \(I = \int \frac{\cos x - \sin x}{\cos x + \sin x}\,dx\). Put \(\cos x + \sin x = t\) so that \((-\sin x + \cos x)\,dx = dt\). Therefore \(I = \log|\cos x + \sin x| + C_1\).
Putting it in (1): \(\frac{1}{1+\tan x} = \frac{\cos x}{\cos x + \sin x} = \frac{1}{2}\left[1 + \frac{\cos x - \sin x}{\cos x + \sin x}\right]\).
Thus \(\int \frac{dx}{1+\tan x} = \frac{1}{2}\left[x + \log|\cos x + \sin x|\right] + C\).

7.3.2 Integration Using Trigonometric Identities

When the integrand involves some trigonometric functions, we use some known identities to find the integral as illustrated through the following example.

Example 7

Find (i) \(\int \cos^2 x\,dx\) (ii) \(\int 2\cos 3x\,dx\) (iii) \(\int \sin^3 x\,dx\)

Solution

(i) Recall the identity \(\cos 2x = 2\cos^2 x - 1\), which gives \(\cos^2 x = \frac{1+\cos 2x}{2}\). \[\int \cos^2 x\,dx = \frac{1}{2}\int(1 + \cos 2x)\,dx = \frac{1}{2}\left(x + \frac{\sin 2x}{2}\right) + C = \frac{x}{2} + \frac{\sin 2x}{4} + C\]

(ii) Recall the identity \(\sin a\cos b = \frac{1}{2}[\sin(a+b) + \sin(a-b)]\). \[\int 2\cos 3x\sin x\,dx = \int [\sin(3x+x) - \sin(3x-x)]\,dx = \int[\sin 4x - \sin 2x]\,dx\] \[= -\frac{\cos 4x}{4} + \frac{\cos 2x}{2} + C\]

(iii) From the identity \(\sin 3x = 3\sin x - 4\sin^3 x\), we find that \(\sin^3 x = \frac{3\sin x - \sin 3x}{4}\). \[\int \sin^3 x\,dx = \frac{1}{4}\int(3\sin x - \sin 3x)\,dx = \frac{1}{4}\left[-3\cos x + \frac{\cos 3x}{3}\right] + C\] \[= -\frac{3}{4}\cos x + \frac{1}{12}\cos 3x + C\]

Alternative for (iii): \(\int \sin^3 x\,dx = \int \sin^2 x\cdot\sin x\,dx = \int (1-\cos^2 x)\sin x\,dx\). Put \(\cos x = t\), \(-\sin x\,dx = dt\): \[= -\int(1-t^2)\,dt = -t + \frac{t^3}{3} + C = -\cos x + \frac{\cos^3 x}{3} + C\]

Interactive: Substitution Checker

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Exercise 7.2

Integrate the functions in Exercises 1 to 37:

1. \(\frac{2x}{1+x^2}\)

Put \(1+x^2 = t\), \(2x\,dx = dt\). \(\int \frac{dt}{t} = \log|t| + C = \log|1+x^2| + C = \log(1+x^2) + C\) (since \(1+x^2 > 0\)).

2. \(\frac{(\log x)^2}{x}\)

Put \(\log x = t\), \(\frac{1}{x}\,dx = dt\). \(\int t^2\,dt = \frac{t^3}{3} + C = \frac{(\log x)^3}{3} + C\).

3. \(\frac{1}{x + x\log x}\)

\(\frac{1}{x(1+\log x)}\). Put \(1+\log x = t\), \(\frac{1}{x}\,dx = dt\). \(\int \frac{dt}{t} = \log|t| + C = \log|1 + \log x| + C\).

4. \(\sin x\cdot\sin(\cos x)\)

Put \(\cos x = t\), \(-\sin x\,dx = dt\). \(-\int \sin t\,dt = \cos t + C = \cos(\cos x) + C\).

5. \(\sin(ax+b)\cos(ax+b)\)

Using \(2\sin\theta\cos\theta = \sin 2\theta\): \(\frac{1}{2}\sin 2(ax+b)\). \(\int \frac{1}{2}\sin(2ax + 2b)\,dx = -\frac{1}{4a}\cos(2ax + 2b) + C\).

6. \(\sqrt{ax+b}\)

Put \(ax+b = t\), \(a\,dx = dt\). \(\frac{1}{a}\int t^{1/2}\,dt = \frac{1}{a}\cdot\frac{2}{3}t^{3/2} + C = \frac{2}{3a}(ax+b)^{3/2} + C\).

7. \(x\sqrt{x+2}\)

Put \(x+2 = t\), \(dx = dt\), \(x = t-2\). \(\int (t-2)\sqrt{t}\,dt = \int (t^{3/2} - 2t^{1/2})\,dt = \frac{2}{5}t^{5/2} - \frac{4}{3}t^{3/2} + C = \frac{2}{5}(x+2)^{5/2} - \frac{4}{3}(x+2)^{3/2} + C\).

8. \(x\sqrt{1+2x^2}\)

Put \(1+2x^2 = t\), \(4x\,dx = dt\). \(\frac{1}{4}\int \sqrt{t}\,dt = \frac{1}{4}\cdot\frac{2}{3}t^{3/2} + C = \frac{1}{6}(1+2x^2)^{3/2} + C\).

9. \((4x+2)\sqrt{x^2+x+1}\)

Note \(\frac{d}{dx}(x^2+x+1) = 2x+1\). So \(4x+2 = 2(2x+1)\). Put \(x^2+x+1 = t\), \((2x+1)\,dx = dt\). \(2\int \sqrt{t}\,dt = \frac{4}{3}t^{3/2} + C = \frac{4}{3}(x^2+x+1)^{3/2} + C\).

10. \(\frac{1}{x - \sqrt{x}}\)

\(\frac{1}{\sqrt{x}(\sqrt{x}-1)}\). Put \(\sqrt{x} - 1 = t\), \(\frac{1}{2\sqrt{x}}\,dx = dt\). \(2\int \frac{dt}{t} = 2\log|t| + C = 2\log|\sqrt{x}-1| + C\).

11. \(\frac{x}{\sqrt{x+4}},\; x > 0\)

Put \(x+4 = t^2\), \(dx = 2t\,dt\), \(x = t^2 - 4\). \(\int \frac{(t^2-4)}{t}\cdot 2t\,dt = 2\int(t^2-4)\,dt = \frac{2t^3}{3} - 8t + C = \frac{2}{3}(x+4)^{3/2} - 8\sqrt{x+4} + C\).

12. \((x^3 - 1)^{1/3}x^5\)

Write \(x^5 = x^2 \cdot x^3\). Put \(x^3 - 1 = t\), \(3x^2\,dx = dt\), \(x^3 = t+1\). \(\frac{1}{3}\int t^{1/3}(t+1)\,dt = \frac{1}{3}\int(t^{4/3} + t^{1/3})\,dt = \frac{1}{3}\left[\frac{3}{7}t^{7/3} + \frac{3}{4}t^{4/3}\right] + C = \frac{1}{7}(x^3-1)^{7/3} + \frac{1}{4}(x^3-1)^{4/3} + C\).

Exercises 13 to 22 (Trigonometric integrals):

13. \(\sin^2(2x+5)\)

Using \(\sin^2\theta = \frac{1-\cos 2\theta}{2}\): \(\int \frac{1-\cos(4x+10)}{2}\,dx = \frac{x}{2} - \frac{\sin(4x+10)}{8} + C\).

14. \(\sin 3x\cos 4x\)

Using \(2\sin A\cos B = \sin(A+B) + \sin(A-B)\): \(\frac{1}{2}\int[\sin 7x + \sin(-x)]\,dx = \frac{1}{2}\left[-\frac{\cos 7x}{7} + \cos x\right] + C\).

15. \(\cos 2x\cos 4x\cos 6x\)

Using product-to-sum: \(\cos 2x\cos 4x = \frac{1}{2}[\cos 6x + \cos 2x]\). Then multiply by \(\cos 6x\): \(\frac{1}{2}[\cos 6x\cos 6x + \cos 2x\cos 6x] = \frac{1}{2}\left[\frac{1+\cos 12x}{2} + \frac{\cos 8x + \cos 4x}{2}\right]\). Integrating: \[\frac{1}{4}\left[x + \frac{\sin 12x}{12} + \frac{\sin 8x}{8} + \frac{\sin 4x}{4}\right] + C\]

Choose the correct answer in Exercises 23 and 24:

23. \(\int \frac{dx}{x(x^2+1)}\) equals:
(A) \(\log|x| - \frac{1}{2}\log(x^2+1) + C\) (B) \(\log|x| + \frac{1}{2}\log(x^2+1) + C\)
(C) \(-\log|x| + \frac{1}{2}\log(x^2+1) + C\) (D) \(\frac{1}{2}\log|x| + \log(x^2+1) + C\)

\(\frac{1}{x(x^2+1)} = \frac{1}{x} - \frac{x}{x^2+1}\) (partial fractions). Integrating: \(\log|x| - \frac{1}{2}\log(x^2+1) + C\). Answer: (A)

24. \(\int \frac{dx}{\sqrt{\sin^3 x\cos^5 x}}\) — this is an advanced integral covered in later sections.

Activity: Spotting the Substitution

L4 Analyse

Predict: For each integral below, can you identify what substitution to use before computing the answer?

Look at the integrand and ask: "Is there a function whose derivative also appears as a factor?"
Try these: (a) \(\int x e^{x^2}\,dx\) (b) \(\int \frac{\cos x}{1+\sin x}\,dx\) (c) \(\int \frac{2x}{(1+x^2)^2}\,dx\)
Write down your chosen substitution \(t = \ldots\) for each.
Verify by computing the derivative \(dt\) and checking that the integral simplifies.

(a) Substitution: \(t = x^2\), \(dt = 2x\,dx\). Result: \(\frac{1}{2}e^{x^2} + C\).

(b) Substitution: \(t = 1 + \sin x\), \(dt = \cos x\,dx\). Result: \(\log|1+\sin x| + C\).

Competency-Based Questions

Scenario: A chemical reaction has a rate given by \(r(t) = \frac{k}{(1+t)^2}\) mol/s, where \(k = 0.5\) and \(t\) is time in seconds. The total amount of product formed from \(t = 0\) is given by the integral of the rate function.

Q1. Using substitution \(u = 1+t\), the integral \(\int \frac{0.5}{(1+t)^2}\,dt\) equals:

L3 Apply

(a) \(\frac{0.5}{1+t} + C\)
(b) \(-\frac{0.5}{1+t} + C\)
(c) \(0.5\log(1+t) + C\)
(d) \(-0.5\log(1+t) + C\)

Answer: (b). Let \(u = 1+t\), \(du = dt\). \(\int \frac{0.5}{u^2}\,du = 0.5 \cdot \frac{u^{-1}}{-1} + C = -\frac{0.5}{u} + C = -\frac{0.5}{1+t} + C\).

Q2. If initially no product is formed, find the amount of product at \(t = 4\) seconds.

L3 Apply

Answer: Amount \(= -\frac{0.5}{1+t}\Big|_0^4 = -\frac{0.5}{5} - \left(-\frac{0.5}{1}\right) = -0.1 + 0.5 = 0.4\) mol.

Q3. Analyse: As \(t \to \infty\), what is the maximum product that can ever be formed? Explain why the reaction never truly completes.

L4 Analyse

Answer: As \(t \to \infty\): \(-\frac{0.5}{1+t} + 0.5 \to 0 + 0.5 = 0.5\) mol. The maximum product is 0.5 mol. The reaction asymptotically approaches this limit because the rate \(\frac{0.5}{(1+t)^2}\) decreases as \(\frac{1}{t^2}\), which approaches zero but never equals zero. So the reaction perpetually slows down without ever stopping completely.

Q4. A student claims: "If the rate were \(\frac{k}{1+t}\) instead of \(\frac{k}{(1+t)^2}\), the total product would also approach a finite limit." Evaluate this claim.

L5 Evaluate

Answer: The claim is false. \(\int_0^\infty \frac{k}{1+t}\,dt = k\log(1+t)\Big|_0^\infty = \infty\). The integral diverges, so the total product would be unbounded. The key difference is that \(\frac{1}{(1+t)^2}\) decreases fast enough for the integral to converge, while \(\frac{1}{1+t}\) does not.

Assertion–Reason Questions

Assertion (A): \(\int \tan x\,dx = \log|\sec x| + C\).
Reason (R): Substituting \(t = \cos x\) transforms \(\int \frac{\sin x}{\cos x}\,dx\) into \(-\int \frac{dt}{t}\).

(a) Both true, R explains A.

(b) Both true, R doesn't explain A.

(c) A true, R false.

(d) A false, R true.

Answer: (a) — Both true. With \(t = \cos x\): \(-\int \frac{dt}{t} = -\log|t| + C = -\log|\cos x| + C = \log|\sec x| + C\). R directly explains how A is derived.

Assertion (A): \(\int 2x\sin(x^2)\,dx = -\cos(x^2) + C\).
Reason (R): If \(g'(x)\) appears as a factor in the integrand alongside \(f(g(x))\), the substitution \(t = g(x)\) simplifies the integral to \(\int f(t)\,dt\).

(a) Both true, R explains A.

(b) Both true, R doesn't explain A.

(c) A true, R false.

(d) A false, R true.

Answer: (a) — Both true. Here \(g(x) = x^2\), \(g'(x) = 2x\) appears as a factor, and \(f = \sin\). Setting \(t = x^2\): \(\int \sin t\,dt = -\cos t + C = -\cos(x^2) + C\). R precisely describes the method used.

Frequently Asked Questions

What is integration by substitution?

Integration by substitution simplifies integrals by replacing a complex expression with a simpler variable u. If the integrand contains f(g(x)) * g'(x), substitute u = g(x) to simplify.

When should you use substitution in integration?

Use when you spot a function and its derivative within the integrand. Look for patterns like f(g(x))*g'(x) or expressions where a substitution simplifies the form.

What are common trigonometric substitutions?

For sqrt(a^2-x^2), use x = a*sin(theta). For sqrt(a^2+x^2), use x = a*tan(theta). For sqrt(x^2-a^2), use x = a*sec(theta).

How to integrate tan(x) and cot(x)?

Integral of tan(x) = log|sec(x)| + C by substituting u = cos(x). Integral of cot(x) = log|sin(x)| + C. These are standard results from substitution.

What are special integral forms using substitution?

Important forms: integral of f'(x)/f(x) = log|f(x)| + C, integral of [f(x)]^n * f'(x) = [f(x)]^(n+1)/(n+1) + C. Also integrals with completing the square.

Frequently Asked Questions — Integrals

What is Integration by Substitution in NCERT Class 12 Mathematics?

Integration by Substitution is a key concept covered in NCERT Class 12 Mathematics, Chapter 7: Integrals. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.

How do I solve problems on Integration by Substitution step by step?

To solve problems on Integration by Substitution, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 12 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.

What are the most important formulas for Chapter 7: Integrals?

The essential formulas of Chapter 7 (Integrals) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.

Is Integration by Substitution important for the Class 12 board exam?

Integration by Substitution is part of the NCERT Class 12 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.

What mistakes should students avoid in Integration by Substitution?

Common mistakes in Integration by Substitution include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.

Where can I find more NCERT practice questions on Integration by Substitution?

End-of-chapter NCERT exercises for Integration by Substitution cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 7, and solve at least one previous-year board paper to consolidate your understanding.

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