Anand Classes is pleased to provide a free downloadable PDF of the NCERT Solutions for Class 12 Maths Chapter 7 – Integrals, Exercise 7.3 (Set-3). This set is thoughtfully crafted to help students master trigonometric-integral problems such as those involving powers of sine and cosine, products like sinA cosB, and repeated products of cosines—topics highlighted in standard solution collections for Exercise 7.3. The detailed, step-by-step solutions adhere strictly to the latest CBSE/NCERT syllabus and are ideal for effective revision and exam readiness. Click the print button to download study material and notes.
NCERT Question 16: Evaluate the Integral
$$\displaystyle \int \tan^4(x)\; dx$$
Solution
Use the identity $\tan^2 x = \sec^2 x – 1$ to write
$$\tan^4 x = (\tan^2 x)^2 = (\sec^2 x – 1)^2 = \sec^4 x – 2\sec^2 x + 1.$$
So
$$\int \tan^4(x)\; dx = \int \sec^4 x\; dx – 2\int \sec^2 x\; dx + \int 1\; dx.$$
Evaluate $\displaystyle\int \sec^4 x\; dx$ by setting $u=\tan x$, $du=\sec^2 x\; dx$:
$$\int \sec^4 x\; dx = \int \sec^2 x(1+\tan^2 x)\; dx $$
$$\int \sec^4 x\; dx = \int (1+u^2)\; du$$
$$\int \sec^4 x\; dx = u + \frac{u^3}{3} + C$$
$$\int \sec^4 x\; dx = \tan x + \frac{1}{3}\tan^3 x + C.$$
Also $\displaystyle\int \sec^2 x\; dx = \tan x$ and $\displaystyle\int 1\; dx = x$.
Therefore
$$
\int \tan^4(x)\; dx
= \Bigl(\tan x + \frac{1}{3}\tan^3 x\Bigr) – 2\tan x + x + C $$
$$\int \tan^4(x)\; dx = x – \tan x + \frac{1}{3}\tan^3 x + C.$$
Final Answer
$$\boxed{\displaystyle \int \tan^4(x)\; dx = x – \tan x + \frac{1}{3}\tan^3 x + C}$$
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NCERT Question 17: Evaluate the Integral
$$\displaystyle \int \frac{\sin^3(x)+\cos^3(x)}{\sin^2(x)\cos^2(x)}\; dx$$
Solution
$$\displaystyle \int \frac{\sin^3(x)+\cos^3(x)}{\sin^2(x)\cos^2(x)}\; dx$$
Simplify the integrand:
$$\frac{\sin^3 x+\cos^3 x}{\sin^2 x\cos^2 x}
=\frac{\sin^3 x}{\sin^2 x\cos^2 x}+\frac{\cos^3 x}{\sin^2 x\cos^2 x}
=\frac{\sin x}{\cos^2 x}+\frac{\cos x}{\sin^2 x}$$
Rewrite using standard trig functions:
$$\frac{\sin x}{\cos^2 x}=\tan x\sec x,\qquad
\frac{\cos x}{\sin^2 x}=\cot x\csc x$$
So the integral becomes
$$\int\bigl(\tan x\sec x+\cot x\csc x\bigr)\; dx
=\int \sec x\tan x\; dx+\int \cot x\csc x\; dx$$
Use known antiderivatives $\dfrac{d}{dx}(\sec x)=\sec x\tan x$
and $\dfrac{d}{dx}(\csc x)=-\csc x\cot x$:
$$\int \sec x\tan x\; dx=\sec x,\qquad
\int \cot x\csc x\; dx=-\csc x$$
Final Answer
$$\boxed{\displaystyle \int \frac{\sin^3(x)+\cos^3(x)}{\sin^2(x)\cos^2(x)}\; dx = \sec x – \csc x + C}$$
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NCERT Question 18: Evaluate the Integral
$$\displaystyle \int \frac{\cos(2x) + 2\sin^2(x)}{\cos^2(x)}\; dx$$
Solution
$$\displaystyle \int \frac{\cos(2x) + 2\sin^2(x)}{\cos^2(x)}\; dx$$
First, expand the numerator using the identity
$$\cos(2x) = \cos^2(x) – \sin^2(x)$$
So,
$$\cos(2x) + 2\sin^2(x) = \cos^2(x) – \sin^2(x) + 2\sin^2(x) = \cos^2(x) + \sin^2(x) = 1$$
Hence the integrand simplifies to
$$\frac{1}{\cos^2(x)} = \sec^2(x)$$
Now integrate:
$$\int \sec^2(x)\; dx = \tan(x) + C$$
Final Answer
$$\boxed{\displaystyle \int \frac{\cos(2x) + 2\sin^2(x)}{\cos^2(x)}\; dx = \tan(x) + C}$$
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NCERT Question 19: Evaluate the Integral
$$\displaystyle \int \frac{1}{\sin x\cos^3 x}\; dx$$
Solution
$$\displaystyle \int \frac{1}{\sin x\cos^3 x}\; dx$$
Note that $\csc x=\dfrac{\sec x}{\tan x}$, so
$$
\frac{1}{\sin x\cos^3 x}=\sec^3 x\csc x=\sec^4 x\cdot\frac{1}{\tan x}.
$$
Using $u=\tan x$ and $du=\sec^2 x\; dx$ we get
$$
\sec^4 x\cdot\frac{1}{\tan x}\; dx
=\frac{\sec^2 x}{\tan x}\cdot\sec^2 x\; dx
=\frac{1+u^2}{u}\; du
=\left(\frac{1}{u}+u\right)\; du.
$$
Integrate:
$$
\int\left(\frac{1}{u}+u\right)\; du
=\ln|u|+\frac{u^2}{2}+C.
$$
Substitute back $u=\tan x$:
$$
\boxed{\displaystyle \int \frac{1}{\sin x\cos^3 x}\; dx
=\ln\bigl|\tan x\bigr|+\frac{1}{2}\tan^2 x + C.}
$$
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NCERT Question 20: Evaluate the Integral
$$\displaystyle \int \frac{\cos(2x)}{\bigl(\cos x+\sin x\bigr)^2}\; dx$$
Solution
$$\displaystyle \int \frac{\cos(2x)}{\bigl(\cos x+\sin x\bigr)^2}\; dx$$
Use the identity
$$\cos(2x)=\cos^2 x-\sin^2 x=(\cos x+\sin x)(\cos x-\sin x).$$
Hence the integrand simplifies:
$$\frac{\cos(2x)}{(\cos x+\sin x)^2}=\frac{(\cos x+\sin x)(\cos x-\sin x)}{(\cos x+\sin x)^2}
=\frac{\cos x-\sin x}{\cos x+\sin x}.$$
Let
$$u=\cos x+\sin x\quad\Rightarrow\quad du=(\cos x-\sin x)\; dx.$$
The integral becomes
$$\int \frac{du}{u}=\ln\bigl|u\bigr|+C.$$
Substituting back $u=\cos x+\sin x$ gives the final result.
Final Answer
$$\boxed{\displaystyle \int \frac{\cos(2x)}{\bigl(\cos x+\sin x\bigr)^2}\; dx = \ln\bigl|\cos x+\sin x\bigr| + C}$$
NCERT Question 21: Evaluate the Integral
$$\displaystyle \int \sin^{-1}(\cos x)\; dx$$
Solution
$$\displaystyle \int \sin^{-1}(\cos x)\; dx$$
Let
$$I=\int \sin^{-1}(\cos x)\; dx.$$
We know that
$$\cos x=\sin\left(\frac{\pi}{2}-x\right),$$
so
$$\sin^{-1}(\cos x)=\sin^{-1}\left[\sin\left(\frac{\pi}{2}-x\right)\right]=\frac{\pi}{2}-x.$$
Therefore,
$$I=\int\left(\frac{\pi}{2}-x\right)\; dx.$$
Integrate termwise:
$$I=\frac{\pi x}{2}-\frac{x^2}{2}+C.$$
Final Answer
$$\boxed{\displaystyle \int \sin^{-1}(\cos x)\; dx=\frac{\pi x}{2}-\frac{x^2}{2}+C}$$
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NCERT Question 22: Evaluate the Integral
$$\displaystyle \int \frac{dx}{\cos(x-a)\cos(x-b)}$$
Solution
$$\displaystyle \int \frac{dx}{\cos(x-a)\cos(x-b)}$$
Use the identity
$$\tan(x-b)-\tan(x-a)=\frac{\sin\bigl((x-b)-(x-a)\bigr)}{\cos(x-a)\cos(x-b)}=\frac{\sin(a-b)}{\cos(x-a)\cos(x-b)}$$
Hence
$$\frac{1}{\cos(x-a)\cos(x-b)}=\frac{\tan(x-b)-\tan(x-a)}{\sin(a-b)}.$$
So
$$\int \frac{dx}{\cos(x-a)\cos(x-b)}=\frac{1}{\sin(a-b)}\int\bigl(\tan(x-b)-\tan(x-a)\bigr)\; dx $$
$$\int \frac{dx}{\cos(x-a)\cos(x-b)}=\frac{1}{\sin(a-b)}\Bigl(-\ln\bigl|\cos(x-b)\bigr|+\ln\bigl|\cos(x-a)\bigr|\Bigr)+C$$
Final Answer
$$\boxed{\displaystyle \int \frac{dx}{\cos(x-a)\cos(x-b)}
=\frac{1}{\sin(a-b)}\ln\left|\frac{\cos(x-a)}{\cos(x-b)}\right|+C}$$
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NCERT Question 23: Evaluate the Integral
$$\displaystyle \int \frac{\sin^2 x – \cos^2 x}{\sin^2 x\;\cos^2 x}\; dx$$
Options:
(A) $\tan x + \cot x + C$
(B) $\tan x + \csc x + C$
(C) $-\tan x + \cot x + C$
(D) $\tan x + \sec x + C$
Solution
$$\displaystyle \int \frac{\sin^2 x – \cos^2 x}{\sin^2 x\;\cos^2 x}\; dx$$
We have
$$\frac{\sin^2 x – \cos^2 x}{\sin^2 x\;\cos^2 x}
= \frac{\sin^2 x}{\sin^2 x\;\cos^2 x} – \frac{\cos^2 x}{\sin^2 x\;\cos^2 x}
= \frac{1}{\cos^2 x} – \frac{1}{\sin^2 x}.$$
That is,
$$\frac{\sin^2 x – \cos^2 x}{\sin^2 x\;\cos^2 x}
= \sec^2 x – \csc^2 x.$$
Now integrate each term:
$$\int (\sec^2 x – \csc^2 x)\; dx
= \int \sec^2 x\; dx – \int \csc^2 x\; dx.$$
We know that
$$\int \sec^2 x\; dx = \tan x,\quad \int \csc^2 x\; dx = -\cot x.$$
Hence,
$$\int (\sec^2 x – \csc^2 x)\; dx = \tan x + \cot x + C.$$
Final Answer
$$\boxed{\displaystyle \tan x + \cot x + C}$$
✅ Correct Option: (A)
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NCERT Question 24: Evaluate the integral
$$\displaystyle \int \frac{e^{x}(1+x)}{\cos^2\bigl(e^{x}x\bigr)}\; dx$$
Options:
(A) $-\cot\bigl(e^{x}x\bigr)+C$
(B) $\tan\bigl(e^{x}x\bigr)+C$
(C) $\tan\bigl(e^{x}\bigr)+C$
(D) $\cot\bigl(e^{x}\bigr)+C$
Solution
$$\displaystyle \int \frac{e^{x}(1+x)}{\cos^2\bigl(e^{x}x\bigr)}\; dx$$
Let
$$u = e^{x}x.$$
Differentiate to get
$$du = \bigl(e^{x}+x e^{x}\bigr)\; dx = e^{x}(1+x)\; dx.$$
Thus the integral becomes
$$\int \frac{e^{x}(1+x)}{\cos^2\bigl(e^{x}x\bigr)}\; dx
= \int \frac{du}{\cos^2 u}
= \int \sec^2 u\; du.$$
Integrate:
$$\int \sec^2 u\; du = \tan u + C.$$
Substitute back $u = e^{x}x$ to obtain
$$\boxed{\displaystyle \tan\bigl(e^{x}x\bigr) + C}$$
Final choice
Correct option: (B).
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