Class 12 Maths NCERT Solutions Chapter 7 Ex 7.2 – Integration by Substitution | Boundless Maths

Ex 7.2 Class 12 Maths NCERT Solutions

Class 12 Maths NCERT Solutions Chapter 7 Ex 7.2 – Integration by Substitution

These Class 12 Maths NCERT Solutions Chapter 7 Ex 7.2 cover all 39 questions — integration by substitution across trigonometric, exponential, logarithmic and algebraic integrands.

This is the longest exercise in the chapter, and for good reason: substitution is the single most-used integration technique you'll rely on going forward, so NCERT gives you extensive practice across every integrand type. The key skill throughout is spotting a function inside the integrand whose derivative also appears elsewhere, usually as a multiplying factor — that's your signal for what to set as t. Questions 1–37 work through this pattern across algebraic, trigonometric, exponential and logarithmic expressions, roughly increasing in difficulty, while the final two are MCQs that test whether you can recognise a ready-made u′(x)/u(x) form at a glance. Take your time here — the pattern-recognition instinct you build in this exercise carries directly into Ex 7.3 onward.

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Class 12 Maths NCERT Solutions Chapter 7 Ex 7.2 — All 39 Questions

Integrate: \dfrac{2x}{1+x^2}

Easy +

Solution

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

The integral becomes \displaystyle\int \dfrac{dt}{t} = \log|t|.

\displaystyle\int \dfrac{2x}{1+x^2}\,dx = \log(1+x^2) + C

Integrate: \dfrac{(\log x)^2}{x}

Easy +

Solution

Put t=\log x, so dt=\dfrac{1}{x}dx.

The integral becomes \displaystyle\int t^2\,dt = \dfrac{t^3}{3}.

\displaystyle\int \dfrac{(\log x)^2}{x}\,dx = \dfrac{(\log x)^3}{3} + C

Integrate: \dfrac{1}{x+x\log x}

Easy +

Solution

Factor the denominator: x+x\log x = x(1+\log x).

Put t=1+\log x, so dt=\dfrac{1}{x}dx. The integral becomes \displaystyle\int \dfrac{dt}{t} = \log|t|.

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

Integrate: \sin x\,\sin(\cos x)

Easy +

Solution

Put t=\cos x, so dt=-\sin x\,dx.

The integral becomes \displaystyle-\int \sin t\,dt = \cos t.

\displaystyle\int \sin x\sin(\cos x)\,dx = \cos(\cos x) + C

Integrate: \sin(ax+b)\cos(ax+b)

Medium +

Solution

Put t=\sin(ax+b), so dt=a\cos(ax+b)\,dx.

The integral becomes \displaystyle\dfrac{1}{a}\int t\,dt = \dfrac{t^2}{2a}.

\displaystyle\int \sin(ax+b)\cos(ax+b)\,dx = \dfrac{\sin^2(ax+b)}{2a} + C

Integrate: \sqrt{ax+b}

Easy +

Solution

Put t=ax+b, so dt=a\,dx.

The integral becomes \displaystyle\dfrac{1}{a}\int \sqrt{t}\,dt = \dfrac{1}{a}\cdot\dfrac23 t^{3/2}.

\displaystyle\int \sqrt{ax+b}\,dx = \dfrac{2}{3a}(ax+b)^{3/2} + C

Integrate: x\sqrt{x+2}

Medium +

Solution

Put t=x+2, so x=t-2 and dx=dt.

The integral becomes \displaystyle\int (t-2)\sqrt{t}\,dt = \int\left(t^{3/2}-2t^{1/2}\right)dt = \dfrac25 t^{5/2} - \dfrac43 t^{3/2}.

\displaystyle\int x\sqrt{x+2}\,dx = \dfrac25(x+2)^{5/2} - \dfrac43(x+2)^{3/2} + C

Integrate: x\sqrt{1+2x^2}

Easy +

Solution

Put t=1+2x^2, so dt=4x\,dx.

The integral becomes \displaystyle\dfrac14\int \sqrt{t}\,dt = \dfrac14\cdot\dfrac23 t^{3/2}.

\displaystyle\int x\sqrt{1+2x^2}\,dx = \dfrac16(1+2x^2)^{3/2} + C

Integrate: (4x+2)\sqrt{x^2+x+1}

Medium +

Solution

Put t=x^2+x+1, so dt=(2x+1)\,dx. Note 4x+2 = 2(2x+1).

The integral becomes \displaystyle 2\int \sqrt{t}\,dt = 2\cdot\dfrac23 t^{3/2}.

\displaystyle\int (4x+2)\sqrt{x^2+x+1}\,dx = \dfrac43(x^2+x+1)^{3/2} + C

Integrate: \dfrac{1}{x-\sqrt{x}}

Medium +

Solution

Factor the denominator: x-\sqrt{x} = \sqrt{x}(\sqrt{x}-1).

Put t=\sqrt{x}, so x=t^2 and dx=2t\,dt.

The integral becomes \displaystyle\int \dfrac{2t\,dt}{t(t-1)} = 2\int \dfrac{dt}{t-1} = 2\log|t-1|.

\displaystyle\int \dfrac{1}{x-\sqrt{x}}\,dx = 2\log|\sqrt{x}-1| + C

Integrate: \dfrac{x}{\sqrt{x+4}}, x>0

Medium +

Solution

Put t=\sqrt{x+4}, so x=t^2-4 and dx=2t\,dt.

The integral becomes \displaystyle\int \dfrac{(t^2-4)}{t}\cdot 2t\,dt = 2\int (t^2-4)\,dt = \dfrac{2t^3}{3} - 8t.

\displaystyle\int \dfrac{x}{\sqrt{x+4}}\,dx = \dfrac23(x+4)^{3/2} - 8\sqrt{x+4} + C

Integrate: (x^3-1)^{1/3}x^5

Hard +

Solution

Put t=x^3-1, so dt=3x^2\,dx and x^3=t+1.

Write x^5\,dx = x^3\cdot x^2\,dx = (t+1)\cdot\dfrac{dt}{3}, so the integral becomes \displaystyle\dfrac13\int t^{1/3}(t+1)\,dt = \dfrac13\int\left(t^{4/3}+t^{1/3}\right)dt.

Integrate: \displaystyle\dfrac13\left[\dfrac37 t^{7/3} + \dfrac34 t^{4/3}\right] = \dfrac17 t^{7/3} + \dfrac14 t^{4/3}.

\displaystyle\int (x^3-1)^{1/3}x^5\,dx = \dfrac17(x^3-1)^{7/3} + \dfrac14(x^3-1)^{4/3} + C

Integrate: \dfrac{x^2}{(2+3x^3)^3}

Medium +

Solution

Put t=2+3x^3, so dt=9x^2\,dx.

The integral becomes \displaystyle\dfrac19\int t^{-3}\,dt = \dfrac19\cdot\dfrac{t^{-2}}{-2} = -\dfrac{1}{18t^2}.

\displaystyle\int \dfrac{x^2}{(2+3x^3)^3}\,dx = -\dfrac{1}{18(2+3x^3)^2} + C

Integrate: \dfrac{1}{x(\log x)^m}, x>0,\,m\neq 1

Medium +

Solution

Put t=\log x, so dt=\dfrac{1}{x}dx.

The integral becomes \displaystyle\int t^{-m}\,dt = \dfrac{t^{1-m}}{1-m} (valid since m\neq1).

\displaystyle\int \dfrac{1}{x(\log x)^m}\,dx = \dfrac{(\log x)^{1-m}}{1-m} + C

Integrate: \dfrac{x}{9-4x^2}

Easy +

Solution

Put t=9-4x^2, so dt=-8x\,dx.

The integral becomes \displaystyle-\dfrac18\int \dfrac{dt}{t} = -\dfrac18\log|t|.

\displaystyle\int \dfrac{x}{9-4x^2}\,dx = -\dfrac18\log|9-4x^2| + C

Integrate: e^{2x+3}

Easy +

Solution

Put t=2x+3, so dt=2\,dx.

The integral becomes \displaystyle\dfrac12\int e^t\,dt = \dfrac12 e^t.

\displaystyle\int e^{2x+3}\,dx = \dfrac12 e^{2x+3} + C

Integrate: \dfrac{x}{e^{x^2}}

Easy +

Solution

Put t=x^2, so dt=2x\,dx.

The integral becomes \displaystyle\dfrac12\int e^{-t}\,dt = -\dfrac12 e^{-t}.

\displaystyle\int \dfrac{x}{e^{x^2}}\,dx = -\dfrac12 e^{-x^2} + C

Integrate: \dfrac{e^{\tan^{-1}x}}{1+x^2}

Easy +

Solution

Put t=\tan^{-1}x, so dt=\dfrac{1}{1+x^2}dx.

The integral becomes \displaystyle\int e^t\,dt = e^t.

\displaystyle\int \dfrac{e^{\tan^{-1}x}}{1+x^2}\,dx = e^{\tan^{-1}x} + C

Integrate: \dfrac{e^{2x}-1}{e^{2x}+1}

Medium +

Solution

Divide the numerator and denominator by e^x: \dfrac{e^{2x}-1}{e^{2x}+1} = \dfrac{e^x-e^{-x}}{e^x+e^{-x}}.

Put t=e^x+e^{-x}, so dt=(e^x-e^{-x})\,dx — exactly the numerator.

The integral becomes \displaystyle\int \dfrac{dt}{t} = \log|t|.

\displaystyle\int \dfrac{e^{2x}-1}{e^{2x}+1}\,dx = \log(e^x+e^{-x}) + C

Integrate: \dfrac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}

Medium +

Solution

Put t=e^{2x}+e^{-2x}, so dt=2(e^{2x}-e^{-2x})\,dx.

The integral becomes \displaystyle\dfrac12\int \dfrac{dt}{t} = \dfrac12\log|t|.

\displaystyle\int \dfrac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}\,dx = \dfrac12\log(e^{2x}+e^{-2x}) + C

Integrate: \tan^2(2x-3)

Easy +

Solution

Use the identity \tan^2\theta = \sec^2\theta - 1: \tan^2(2x-3) = \sec^2(2x-3) - 1.

Since \displaystyle\int \sec^2(2x-3)\,dx = \dfrac12\tan(2x-3), subtract x for the -1 term.

\displaystyle\int \tan^2(2x-3)\,dx = \dfrac12\tan(2x-3) - x + C

Integrate: \sec^2(7-4x)

Easy +

Solution

Put t=7-4x, so dt=-4\,dx.

The integral becomes \displaystyle-\dfrac14\int \sec^2 t\,dt = -\dfrac14\tan t.

\displaystyle\int \sec^2(7-4x)\,dx = -\dfrac14\tan(7-4x) + C

Integrate: \dfrac{\sin^{-1}x}{\sqrt{1-x^2}}

Medium +

Solution

Put t=\sin^{-1}x, so dt=\dfrac{1}{\sqrt{1-x^2}}dx.

The integral becomes \displaystyle\int t\,dt = \dfrac{t^2}{2}.

\displaystyle\int \dfrac{\sin^{-1}x}{\sqrt{1-x^2}}\,dx = \dfrac{(\sin^{-1}x)^2}{2} + C

Integrate: \dfrac{2\cos x-3\sin x}{6\cos x+4\sin x}

Hard +

Solution

Express the numerator as A(6\cos x+4\sin x)+B\dfrac{d}{dx}(6\cos x+4\sin x), i.e. 2\cos x-3\sin x = A(6\cos x+4\sin x)+B(4\cos x-6\sin x).

Comparing coefficients: 6A+4B=2 and 4A-6B=-3. Solving gives A=0, B=\dfrac12.

So the numerator is exactly \dfrac12 times the derivative of the denominator, so the integral becomes \displaystyle\dfrac12\int \dfrac{d(6\cos x+4\sin x)}{6\cos x+4\sin x} = \dfrac12\log|6\cos x+4\sin x|.

\displaystyle\int \dfrac{2\cos x-3\sin x}{6\cos x+4\sin x}\,dx = \dfrac12\log|6\cos x+4\sin x| + C

Integrate: \dfrac{1}{\cos^2 x\,(1-\tan x)^2}

Medium +

Solution

Put t=\tan x, so dt=\sec^2 x\,dx = \dfrac{dx}{\cos^2 x}.

The integral becomes \displaystyle\int \dfrac{dt}{(1-t)^2}. Since \dfrac{d}{dt}\left[\dfrac{1}{1-t}\right] = \dfrac{1}{(1-t)^2}, this evaluates to \dfrac{1}{1-t}.

\displaystyle\int \dfrac{1}{\cos^2 x(1-\tan x)^2}\,dx = \dfrac{1}{1-\tan x} + C

Integrate: \dfrac{\cos\sqrt{x}}{\sqrt{x}}

Easy +

Solution

Put t=\sqrt{x}, so dt=\dfrac{1}{2\sqrt{x}}dx, i.e. \dfrac{dx}{\sqrt{x}}=2\,dt.

The integral becomes \displaystyle 2\int \cos t\,dt = 2\sin t.

\displaystyle\int \dfrac{\cos\sqrt{x}}{\sqrt{x}}\,dx = 2\sin\sqrt{x} + C

Integrate: \sqrt{\sin 2x}\,\cos 2x

Medium +

Solution

Put t=\sin 2x, so dt=2\cos 2x\,dx.

The integral becomes \displaystyle\dfrac12\int \sqrt{t}\,dt = \dfrac12\cdot\dfrac23 t^{3/2}.

\displaystyle\int \sqrt{\sin 2x}\,\cos 2x\,dx = \dfrac13(\sin 2x)^{3/2} + C

Integrate: \dfrac{\cos x}{\sqrt{1+\sin x}}

Easy +

Solution

Put t=1+\sin x, so dt=\cos x\,dx.

The integral becomes \displaystyle\int t^{-1/2}\,dt = 2\sqrt{t}.

\displaystyle\int \dfrac{\cos x}{\sqrt{1+\sin x}}\,dx = 2\sqrt{1+\sin x} + C

Integrate: \cot x\,\log(\sin x)

Easy +

Solution

Put t=\log(\sin x), so dt=\dfrac{\cos x}{\sin x}dx = \cot x\,dx.

The integral becomes \displaystyle\int t\,dt = \dfrac{t^2}{2}.

\displaystyle\int \cot x\log(\sin x)\,dx = \dfrac{[\log(\sin x)]^2}{2} + C

Integrate: \dfrac{\sin x}{1+\cos x}

Easy +

Solution

Put t=1+\cos x, so dt=-\sin x\,dx.

The integral becomes \displaystyle-\int \dfrac{dt}{t} = -\log|t|.

\displaystyle\int \dfrac{\sin x}{1+\cos x}\,dx = -\log(1+\cos x) + C

Integrate: \dfrac{\sin x}{(1+\cos x)^2}

Easy +

Solution

Put t=1+\cos x, so dt=-\sin x\,dx.

The integral becomes \displaystyle-\int t^{-2}\,dt = \dfrac{1}{t}.

\displaystyle\int \dfrac{\sin x}{(1+\cos x)^2}\,dx = \dfrac{1}{1+\cos x} + C

Integrate: \dfrac{1}{1+\cot x}

Hard +

Solution

Rewrite: \dfrac{1}{1+\cot x} = \dfrac{\sin x}{\sin x+\cos x}.

Express the numerator as A(\sin x+\cos x)+B\dfrac{d}{dx}(\sin x+\cos x): \sin x = A(\sin x+\cos x)+B(\cos x-\sin x).

Comparing coefficients gives A-B=1 and A+B=0, so A=\dfrac12, B=-\dfrac12.

So the integral is \displaystyle\int \left[A + B\cdot\dfrac{\cos x-\sin x}{\sin x+\cos x}\right]dx = Ax + B\log|\sin x+\cos x|.

\displaystyle\int \dfrac{1}{1+\cot x}\,dx = \dfrac{x}{2} - \dfrac12\log|\sin x+\cos x| + C

Integrate: \dfrac{1}{1-\tan x}

Hard +

Solution

Rewrite: \dfrac{1}{1-\tan x} = \dfrac{\cos x}{\cos x-\sin x}.

Express the numerator as A(\cos x-\sin x)+B\dfrac{d}{dx}(\cos x-\sin x): \cos x = A(\cos x-\sin x)+B(-\sin x-\cos x).

Comparing coefficients gives A-B=1 and -A-B=0, so A=\dfrac12, B=-\dfrac12.

So the integral is \displaystyle\int \left[A + B\cdot\dfrac{-\sin x-\cos x}{\cos x-\sin x}\right]dx = Ax + B\log|\cos x-\sin x|.

\displaystyle\int \dfrac{1}{1-\tan x}\,dx = \dfrac{x}{2} - \dfrac12\log|\cos x-\sin x| + C

Integrate: \dfrac{\sqrt{\tan x}}{\sin x\cos x}

Medium +

Solution

Write \sin x\cos x = \cos^2 x\tan x, so the integrand becomes \dfrac{\sqrt{\tan x}}{\cos^2 x\tan x} = \dfrac{\sec^2 x}{\sqrt{\tan x}}.

Put t=\tan x, so dt=\sec^2 x\,dx. The integral becomes \displaystyle\int t^{-1/2}\,dt = 2\sqrt{t}.

\displaystyle\int \dfrac{\sqrt{\tan x}}{\sin x\cos x}\,dx = 2\sqrt{\tan x} + C

Integrate: \dfrac{(1+\log x)^2}{x}

Easy +

Solution

Put t=1+\log x, so dt=\dfrac{1}{x}dx.

The integral becomes \displaystyle\int t^2\,dt = \dfrac{t^3}{3}.

\displaystyle\int \dfrac{(1+\log x)^2}{x}\,dx = \dfrac{(1+\log x)^3}{3} + C

Integrate: \dfrac{(x+1)(x+\log x)^2}{x}

Medium +

Solution

Notice \dfrac{d}{dx}(x+\log x) = 1+\dfrac{1}{x} = \dfrac{x+1}{x}, which is exactly the leading factor in the integrand.

Put t=x+\log x, so dt=\dfrac{x+1}{x}\,dx. The integral becomes \displaystyle\int t^2\,dt = \dfrac{t^3}{3}.

\displaystyle\int \dfrac{(x+1)(x+\log x)^2}{x}\,dx = \dfrac{(x+\log x)^3}{3} + C

Integrate: \dfrac{x^3\sin(\tan^{-1}x^4)}{1+x^8}

Hard +

Solution

Put t=\tan^{-1}(x^4), so dt=\dfrac{4x^3}{1+x^8}dx, i.e. \dfrac{x^3\,dx}{1+x^8}=\dfrac{dt}{4}.

The integral becomes \displaystyle\dfrac14\int \sin t\,dt = -\dfrac14\cos t.

\displaystyle\int \dfrac{x^3\sin(\tan^{-1}x^4)}{1+x^8}\,dx = -\dfrac14\cos(\tan^{-1}x^4) + C

MCQ. \displaystyle\int \dfrac{10x^9+10^x\log_e 10}{x^{10}+10^x}\,dx equals: (A) 10^x-x^{10}+C (B) 10^x+x^{10}+C (C) (10^x-x^{10})^{-1}+C (D) \log(10^x+x^{10})+C

Medium +

Solution

Notice that \dfrac{d}{dx}(x^{10}+10^x) = 10x^9 + 10^x\log_e 10 — exactly the numerator.

So the integrand has the form \dfrac{u'(x)}{u(x)}, whose integral is \log|u(x)|.

Answer: (D) \log(10^x+x^{10})+C

MCQ. \displaystyle\int \dfrac{dx}{\sin^2 x\cos^2 x} equals: (A) \tan x+\cot x+C (B) \tan x-\cot x+C (C) \tan x\cot x+C (D) \tan x-\cot 2x+C

Medium +

Solution

Use 1=\sin^2 x+\cos^2 x in the numerator: \dfrac{1}{\sin^2 x\cos^2 x} = \dfrac{\sin^2 x+\cos^2 x}{\sin^2 x\cos^2 x} = \dfrac{1}{\cos^2 x}+\dfrac{1}{\sin^2 x} = \sec^2 x+\text{cosec}^2x.

Integrate term by term: \displaystyle\int \sec^2 x\,dx = \tan x and \displaystyle\int \text{cosec}^2x\,dx = -\cot x.

Answer: (B) \tan x-\cot x+C

Common Questions

Class 12 Maths NCERT Solutions Chapter 7 Ex 7.2 — FAQs

How many questions are there in Class 12 Maths Chapter 7 Ex 7.2?

Ex 7.2 has 39 questions in total — 37 integrals to be solved using the method of substitution, followed by 2 MCQs.

How do I know what substitution to use in an integration problem?

Look for a part of the integrand whose derivative also appears elsewhere in the expression, usually as a multiplying factor. That part is a good candidate for the substitution t. For example, if the integrand contains sin x raised to some power along with cos x, substituting t = sin x works well because d(sin x) = cos x dx.

Where can I check these solutions against the official NCERT textbook?

The official NCERT Class 12 Maths textbook is available on the NCERT website, and every question in Ex 7.2 on this page follows the same numbering as the textbook.

Class 12 Maths NCERT Solutions Chapter 7 Ex 7.2 – Integration by Substitution

Class 12 Maths NCERT Solutions Chapter 7 Ex 7.2 — All 39 Questions

Integrate: \dfrac{2x}{1+x^2}

Integrate: \dfrac{(\log x)^2}{x}

Integrate: \dfrac{1}{x+x\log x}

Integrate: \sin x\,\sin(\cos x)

Integrate: \sin(ax+b)\cos(ax+b)

Integrate: \sqrt{ax+b}

Integrate: x\sqrt{x+2}

Integrate: x\sqrt{1+2x^2}

Integrate: (4x+2)\sqrt{x^2+x+1}

Integrate: \dfrac{1}{x-\sqrt{x}}

Integrate: \dfrac{x}{\sqrt{x+4}}, x>0

Integrate: (x^3-1)^{1/3}x^5

Integrate: \dfrac{x^2}{(2+3x^3)^3}

Integrate: \dfrac{1}{x(\log x)^m}, x>0,\,m\neq 1

Integrate: \dfrac{x}{9-4x^2}

Integrate: e^{2x+3}

Integrate: \dfrac{x}{e^{x^2}}

Integrate: \dfrac{e^{\tan^{-1}x}}{1+x^2}

Integrate: \dfrac{e^{2x}-1}{e^{2x}+1}

Integrate: \dfrac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}

Integrate: \tan^2(2x-3)

Integrate: \sec^2(7-4x)

Integrate: \dfrac{\sin^{-1}x}{\sqrt{1-x^2}}

Integrate: \dfrac{2\cos x-3\sin x}{6\cos x+4\sin x}

Integrate: \dfrac{1}{\cos^2 x\,(1-\tan x)^2}

Integrate: \dfrac{\cos\sqrt{x}}{\sqrt{x}}

Integrate: \sqrt{\sin 2x}\,\cos 2x

Integrate: \dfrac{\cos x}{\sqrt{1+\sin x}}

Integrate: \cot x\,\log(\sin x)

Integrate: \dfrac{\sin x}{1+\cos x}

Integrate: \dfrac{\sin x}{(1+\cos x)^2}

Integrate: \dfrac{1}{1+\cot x}

Integrate: \dfrac{1}{1-\tan x}

Integrate: \dfrac{\sqrt{\tan x}}{\sin x\cos x}

Integrate: \dfrac{(1+\log x)^2}{x}

Integrate: \dfrac{(x+1)(x+\log x)^2}{x}

Integrate: \dfrac{x^3\sin(\tan^{-1}x^4)}{1+x^8}

MCQ. \displaystyle\int \dfrac{10x^9+10^x\log_e 10}{x^{10}+10^x}\,dx equals: (A) 10^x-x^{10}+C (B) 10^x+x^{10}+C (C) (10^x-x^{10})^{-1}+C (D) \log(10^x+x^{10})+C

MCQ. \displaystyle\int \dfrac{dx}{\sin^2 x\cos^2 x} equals: (A) \tan x+\cot x+C (B) \tan x-\cot x+C (C) \tan x\cot x+C (D) \tan x-\cot 2x+C

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Class 12 Maths NCERT Solutions Chapter 7 Ex 7.2 — FAQs

How many questions are there in Class 12 Maths Chapter 7 Ex 7.2?

How do I know what substitution to use in an integration problem?

Where can I check these solutions against the official NCERT textbook?

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