Anand Classes offers a free downloadable PDF of NCERT Solutions for Class 12 Maths Chapter 7 — Integrals, Exercise 7.2 (Set-3), expertly prepared to expand your understanding of integration by substitution and other methods covered in the chapter. The comprehensive material aligns with the latest CBSE / NCERT curriculum and is ideal for thorough revision and enhanced performance in board and competitive exams. Click the print button to download study material and notes.
NCERT Question 21: Find the Integral
$$\int \tan^2(2x-3)\;dx$$
Solution:
$$\int \tan^2(2x-3)\;dx$$
We know that
$$\tan^2(2x-3)=\sec^2(2x-3)-1$$
Let $2x-3=t$
Then $dt=2dx$ or $dx=\frac{1}{2}dt$
Now,
$$
\int \tan^2(2x-3)\;dx
= \frac{1}{2}\int [\sec^2t-1]\;dt
$$
Integrate each term:
$$
= \frac{1}{2}[\tan t – t] + C
$$
Substitute back $t=2x-3$:
$$
\boxed{\frac{1}{2}[\tan(2x-3)-(2x-3)] + C}
$$
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NCERT Question 22: Find the Integral
$$\int \sec^2(7 – 4x) \; dx$$
Solution :
$$\int \sec^2(7 – 4x) \; dx$$
Let $7 – 4x = t$
Then, $dt = -4\;dx$
So,
$$dx = \frac{-1}{4}\;dt$$
Substitute in the integral:
$$\int \sec^2(7 – 4x)\;dx = -\frac{1}{4}\int \sec^2 t \; dt$$
Now integrate:
$$-\frac{1}{4}\int \sec^2 t \;dt = -\frac{1}{4}\tan t + C$$
Substitute back $t = 7 – 4x$:
$$-\frac{1}{4}\tan(7 – 4x) + C$$
Final Answer:
$$\boxed{-\frac{1}{4}\tan(7 – 4x) + C}$$
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NCERT Question 23: Find the Integral
$$\int \frac{\sin^{-1}x}{\sqrt{1 – x^2}}\; dx$$
Solution :
$$\int \frac{\sin^{-1}x}{\sqrt{1 – x^2}}\; dx$$
Let $\sin^{-1}x = t$, then, $\dfrac{1}{\sqrt{1 – x^2}}\;dx = dt$
So,
$$dx = \sqrt{1 – x^2}\; dt$$
Substitute in the integral:
$$\int \frac{\sin^{-1}x}{\sqrt{1 – x^2}}\; dx = \int t \;dt$$
Now integrate:
$$\int t \;dt = \frac{t^2}{2} + C$$
Substitute back $t = \sin^{-1}x$:
$$\frac{(\sin^{-1}x)^2}{2} + C$$
Final Answer:
$$\boxed{\frac{1}{2}(\sin^{-1}x)^2 + C}$$
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NCERT Question 24 : Evaluate the integral
$$\int \frac{2\cos x – 3\sin x}{6\cos x + 4\sin x}dx$$
Solution :
$$\int \frac{2\cos x – 3\sin x}{6\cos x + 4\sin x}dx$$
First observe the denominator can be factored:
$$6\cos x + 4\sin x = 2(3\cos x + 2\sin x)$$
So the integrand becomes
$$\frac{2\cos x – 3\sin x}{6\cos x + 4\sin x}
= \frac{2\cos x – 3\sin x}{2(3\cos x + 2\sin x)}$$
Put
$$t = 3\cos x + 2\sin x.$$
Differentiate to get
$$dt = (-3\sin x + 2\cos x),dx = (2\cos x – 3\sin x)\;dx.$$
Therefore the integral transforms to
$$\int \frac{2\cos x – 3\sin x}{2(3\cos x + 2\sin x)}\;dx
= \frac{1}{2}\int \frac{dt}{t}.$$
Integrating,
$$\frac{1}{2}\int \frac{dt}{t} = \frac{1}{2}\ln|t| + C.$$
Substituting back $t = 3\cos x + 2\sin x$ (equivalently $2\sin x + 3\cos x$),
$$\frac{1}{2}\ln|t| + C = \frac{1}{2}\ln!\big|2\sin x + 3\cos x\big| + C.$$
Final Answer
$$\boxed{\;\frac{1}{2}\ln\big|2\sin x + 3\cos x\big| + C\; }$$
NCERT Question 25: Evaluate the integral
$$\int \frac{1}{\cos^2 x(1 – \tan x)^2}\;dx$$
Solution :
$$\int \frac{1}{\cos^2 x(1 – \tan x)^2}\;dx$$
$$\int \frac{\sec^2 x}{(1 – \tan x)^2}\;dx$$
Let
$$t = 1 – \tan x$$
Then, differentiating both sides gives
$$dt = -\sec^2 x\;dx$$
Substituting these in the integral,
$$\int \frac{\sec^2 x}{(1 – \tan x)^2}\;dx = \int \frac{-dt}{t^2}$$
Simplify,
$$\int \frac{-dt}{t^2} = -\int t^{-2}\;dt$$
Integrating,
$$-\int t^{-2}\;dt = -(-t^{-1}) + C = \frac{1}{t} + C$$
Substitute back $t = 1 – \tan x$,
$$\frac{1}{t} + C = \frac{1}{1 – \tan x} + C$$
Final Answer
$$\boxed{\;\frac{1}{1 – \tan x} + C\;}$$
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NCERT Question 26: Evaluate the integral
$$\int \frac{\cos(\sqrt{x})}{\sqrt{x}}\;dx$$
Solution :
$$\int \frac{\cos(\sqrt{x})}{\sqrt{x}}\;dx$$
Let
$$\sqrt{x} = t$$
Then differentiating,
$$\frac{1}{2\sqrt{x}}\;dx = dt \quad \Rightarrow \quad dx = 2t\;dt$$
Substitute in the integral:
$$\int \frac{\cos(\sqrt{x})}{\sqrt{x}}\;dx = \int \frac{\cos(t)}{t} \times 2t\;dt = 2\int \cos t\;dt$$
Integrating,
$$2\int \cos t\;dt = 2\sin t + C$$
Substitute back $t = \sqrt{x}$:
$$2\sin t + C = 2\sin(\sqrt{x}) + C$$
Final Answer
$$\boxed{\;2\sin(\sqrt{x}\;) + C}$$
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NCERT Question 27: Evaluate the integral
$$\int \sqrt{\sin 2x}\;\cos 2x\;dx$$
Solution :
$$\int \sqrt{\sin 2x}\;\cos 2x\;dx$$
Let
$$\sin 2x = t$$
Then differentiating,
$$2\cos 2x\;dx = dt \quad \Rightarrow \quad \cos 2x\;dx = \frac{1}{2}dt$$
Substitute in the integral:
$$\int \sqrt{\sin 2x}\;\cos 2x\;dx = \int \sqrt{t} \times \frac{1}{2}\;dt = \frac{1}{2}\int t^{\frac{1}{2}}\;dt$$
Integrating,
$$\frac{1}{2}\int t^{\frac{1}{2}}\;dt = \frac{1}{2} \times \frac{t^{\frac{3}{2}}}{\frac{3}{2}} + C = \frac{1}{3}t^{\frac{3}{2}} + C$$
Substitute back $t = \sin 2x$:
$$\frac{1}{3}t^{\frac{3}{2}} + C = \frac{1}{3}(\sin 2x)^{\frac{3}{2}} + C$$
Final Answer
$$\boxed{\;\frac{1}{3}(\sin 2x)^{\frac{3}{2}} + C\;}$$
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NCERT Question 28: Evaluate the integral
$$\int \frac{\cos x}{\sqrt{1 + \sin x}}\;dx$$
Solution :
$$\int \frac{\cos x}{\sqrt{1 + \sin x}}\;dx$$
Let
$$1 + \sin x = t$$
Then differentiating,
$$\cos x\;dx = dt$$
Substitute in the integral:
$$\int \frac{\cos x}{\sqrt{1 + \sin x}}\;dx = \int \frac{dt}{\sqrt{t}}$$
Simplify and integrate:
$$\int \frac{dt}{\sqrt{t}} = \int t^{-\frac{1}{2}}\;dt = \frac{t^{\frac{1}{2}}}{\frac{1}{2}} + C = 2t^{\frac{1}{2}} + C$$
Substitute back $t = 1 + \sin x$:
$$2t^{\frac{1}{2}} + C = 2\sqrt{1 + \sin x} + C$$
Final Answer
$$\boxed{\;2\sqrt{1 + \sin x} + C\;}$$
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NCERT Question 29: Evaluate the integral
$$\int \cot x \;\log(\sin x)\;dx$$
Solution :
$$\int \cot x \;\log(\sin x)\;dx$$
Let
$$\log(\sin x) = t$$
Then differentiating,
$$\frac{1}{\sin x}\cos x\;dx = dt \quad \Rightarrow \quad \cot x\;dx = dt$$
Substitute in the integral:
$$\int \cot x \;\log(\sin x)\;dx = \int t\;dt$$
Integrate:
$$\int t\;dt = \frac{t^2}{2} + C$$
Substitute back $t = \log(\sin x)$:
$$\frac{t^2}{2} + C = \frac{(\log(\sin x))^2}{2} + C$$
Final Answer
$$\boxed{\;\frac{(\log(\sin x))^2}{2} + C\;}$$
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NCERT Question 30: Evaluate the integral
$$\int \frac{\sin x}{1 + \cos x}\;dx$$
Solution :
$$\int \frac{\sin x}{1 + \cos x}\;dx$$
Let
$$1 + \cos x = t$$
Then differentiating,
$$-\sin x\;dx = dt \quad \Rightarrow \quad \sin x\;dx = -dt$$
Substitute in the integral:
$$\int \frac{\sin x}{1 + \cos x}\;dx = \int \frac{-dt}{t} = -\int \frac{dt}{t}$$
Integrate:
$$-\int \frac{dt}{t} = -\log|t| + C$$
Substitute back $t = 1 + \cos x$:
$$-\log|t| + C = -\log|1 + \cos x| + C$$
Final Answer
$$\boxed{\;-\log|1 + \cos x| + C\;}$$
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