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 sin⁡A cos⁡B, 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.

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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

$$\boxed{\displaystyle \int \frac{\cos(2x)}{\bigl(\cos x+\sin x\bigr)^2}\; dx = \ln\bigl|\cos x+\sin x\bigr| + C}$$

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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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