Comprehensive solutions for Class 12 Maths (NCERT) Exercise 7.3 on Integrals. Learn to transform complex trigonometric integrals using Product-to-Sum, Power Reduction ($\sin^2 x$, $\cos^4 x$), and other key identities. Covers types like $\int \sin A \cos B dx$ and $\int \frac{\cos x – \sin x}{1 + \sin 2x} dx$.
Table of Contents
This exercise requires transforming the integrand using trigonometric identities before integration, as simple substitution is not sufficient.
Part 1: Using Trigonometric Power Reduction and Product-to-Sum Identities (1-11)
1. $\int \sin^2 (2x + 5) dx$
Use the identity $\sin^2 \theta = \frac{1 – \cos 2\theta}{2}$.
$$\int \frac{1 – \cos[2(2x + 5)]}{2} dx = \frac{1}{2} \int (1 – \cos(4x + 10)) dx$$
$$= \frac{1}{2} \left[ x – \frac{\sin(4x + 10)}{4} \right] + C$$
$$= \mathbf{\frac{x}{2} – \frac{1}{8} \sin(4x + 10) + C}$$
2. $\int \sin 3x \cos 4x dx$
Use the Product-to-Sum identity $\sin A \cos B = \frac{1}{2} [\sin(A+B) + \sin(A-B)]$.
$$\int \frac{1}{2} [\sin(3x + 4x) + \sin(3x – 4x)] dx = \frac{1}{2} \int [\sin 7x + \sin(-x)] dx$$
$$= \frac{1}{2} \int (\sin 7x – \sin x) dx = \frac{1}{2} \left[ -\frac{\cos 7x}{7} – (-\cos x) \right] + C$$
$$= \mathbf{\frac{1}{2} \left(\cos x – \frac{1}{7} \cos 7x\right) + C}$$
3. $\int \cos 2x \cos 4x \cos 6x dx$
Group the first two terms and use $\cos A \cos B = \frac{1}{2} [\cos(A+B) + \cos(A-B)]$:
$$\frac{1}{2} \int [\cos 6x + \cos 2x] \cos 6x dx = \frac{1}{2} \int (\cos^2 6x + \cos 2x \cos 6x) dx$$
Apply identities again:
- $\cos^2 6x = \frac{1 + \cos 12x}{2}$
- $\cos 2x \cos 6x = \frac{1}{2} [\cos 8x + \cos 4x]$$$\frac{1}{2} \int \left[ \frac{1 + \cos 12x}{2} + \frac{1}{2} (\cos 8x + \cos 4x) \right] dx$$$$= \frac{1}{4} \int (1 + \cos 12x + \cos 8x + \cos 4x) dx$$$$= \mathbf{\frac{1}{4} \left[ x + \frac{\sin 12x}{12} + \frac{\sin 8x}{8} + \frac{\sin 4x}{4} \right] + C}$$
4. $\int \sin^3 (2x + 1) dx$
Use the identity $\sin 3\theta = 3 \sin \theta – 4 \sin^3 \theta \implies \sin^3 \theta = \frac{3 \sin \theta – \sin 3\theta}{4}$.
Let $\theta = 2x + 1$.
$$\int \frac{3 \sin(2x + 1) – \sin(6x + 3)}{4} dx$$
$$= \frac{3}{4} \int \sin(2x + 1) dx – \frac{1}{4} \int \sin(6x + 3) dx$$
$$= \frac{3}{4} \left(-\frac{\cos(2x + 1)}{2}\right) – \frac{1}{4} \left(-\frac{\cos(6x + 3)}{6}\right) + C$$
$$= \mathbf{-\frac{3}{8} \cos(2x + 1) + \frac{1}{24} \cos(6x + 3) + C}$$
5. $\int \sin^3 x \cos^3 x dx$
Rewrite the integrand: $\int (\sin x \cos x)^3 dx = \int \left(\frac{1}{2} \sin 2x\right)^3 dx = \frac{1}{8} \int \sin^3 2x dx$.
Now use the identity $\sin^3 \theta = \frac{3 \sin \theta – \sin 3\theta}{4}$ with $\theta = 2x$.
$$\frac{1}{8} \int \frac{3 \sin 2x – \sin 6x}{4} dx = \frac{1}{32} \int (3 \sin 2x – \sin 6x) dx$$
$$= \frac{1}{32} \left[ 3 \left(-\frac{\cos 2x}{2}\right) – \left(-\frac{\cos 6x}{6}\right) \right] + C$$
$$= \mathbf{\frac{1}{32} \left(\frac{1}{6} \cos 6x – \frac{3}{2} \cos 2x\right) + C}$$
6. $\int \sin x \sin 2x \sin 3x dx$
Group the first two terms: $\sin x \sin 2x$. Use $\sin A \sin B = \frac{1}{2} [\cos(A-B) – \cos(A+B)]$.
$$\frac{1}{2} \int [\cos(x) – \cos(3x)] \sin 3x dx = \frac{1}{2} \int (\cos x \sin 3x – \cos 3x \sin 3x) dx$$
Apply identities again:
- $\cos x \sin 3x = \sin 3x \cos x = \frac{1}{2} [\sin(4x) + \sin(2x)]$
- $\cos 3x \sin 3x = \frac{1}{2} \sin 6x$$$\frac{1}{2} \int \left[ \frac{1}{2} (\sin 4x + \sin 2x) – \frac{1}{2} \sin 6x \right] dx$$$$= \frac{1}{4} \int (\sin 4x + \sin 2x – \sin 6x) dx$$$$= \frac{1}{4} \left[ -\frac{\cos 4x}{4} – \frac{\cos 2x}{2} – (-\frac{\cos 6x}{6}) \right] + C$$$$= \mathbf{\frac{1}{4} \left(\frac{1}{6} \cos 6x – \frac{1}{4} \cos 4x – \frac{1}{2} \cos 2x\right) + C}$$
7. $\int \sin 4x \sin 8x dx$
Use $\sin A \sin B = \frac{1}{2} [\cos(A-B) – \cos(A+B)]$.
$$\int \frac{1}{2} [\cos(4x – 8x) – \cos(4x + 8x)] dx = \frac{1}{2} \int (\cos(-4x) – \cos 12x) dx$$
$$= \frac{1}{2} \int (\cos 4x – \cos 12x) dx = \frac{1}{2} \left[ \frac{\sin 4x}{4} – \frac{\sin 12x}{12} \right] + C$$
$$= \mathbf{\frac{1}{8} \sin 4x – \frac{1}{24} \sin 12x + C}$$
8. $\int \frac{1 – \cos x}{1 + \cos x} dx$
Use half-angle identities: $1 – \cos x = 2 \sin^2(x/2)$ and $1 + \cos x = 2 \cos^2(x/2)$.
$$\int \frac{2 \sin^2(x/2)}{2 \cos^2(x/2)} dx = \int \tan^2(x/2) dx$$
Use $\tan^2 \theta = \sec^2 \theta – 1$:
$$\int [\sec^2(x/2) – 1] dx = \frac{\tan(x/2)}{1/2} – x + C$$
$$= \mathbf{2 \tan(x/2) – x + C}$$
9. $\int \frac{\cos x}{1 + \cos x} dx$
Use the identity $1 = 1 + \cos x – \cos x$:
$$\int \frac{1 + \cos x – 1}{1 + \cos x} dx = \int \left( 1 – \frac{1}{1 + \cos x} \right) dx$$
Use $1 + \cos x = 2 \cos^2(x/2)$:
$$\int \left( 1 – \frac{1}{2 \cos^2(x/2)} \right) dx = \int \left( 1 – \frac{1}{2} \sec^2(x/2) \right) dx$$
$$= x – \frac{1}{2} \frac{\tan(x/2)}{1/2} + C = \mathbf{x – \tan(x/2) + C}$$
10. $\int \sin^4 x dx$
Use $\sin^2 x = \frac{1 – \cos 2x}{2}$.
$$\int (\sin^2 x)^2 dx = \int \left(\frac{1 – \cos 2x}{2}\right)^2 dx = \frac{1}{4} \int (1 – 2 \cos 2x + \cos^2 2x) dx$$
Use $\cos^2 2x = \frac{1 + \cos 4x}{2}$:
$$\frac{1}{4} \int \left(1 – 2 \cos 2x + \frac{1 + \cos 4x}{2}\right) dx = \frac{1}{4} \int \left(\frac{3}{2} – 2 \cos 2x + \frac{1}{2} \cos 4x\right) dx$$
$$= \frac{1}{8} \int (3 – 4 \cos 2x + \cos 4x) dx$$
$$= \frac{1}{8} \left[ 3x – 4 \frac{\sin 2x}{2} + \frac{\sin 4x}{4} \right] + C$$
$$= \mathbf{\frac{3}{8} x – \frac{1}{4} \sin 2x + \frac{1}{32} \sin 4x + C}$$
11. $\int \cos^4 2x dx$
This is the same as Q10, but with the argument $2x$.
$$\int \cos^4 2x dx = \int (\cos^2 2x)^2 dx = \frac{1}{4} \int (1 + 2 \cos 4x + \cos^2 4x) dx$$
Use $\cos^2 4x = \frac{1 + \cos 8x}{2}$:
$$\frac{1}{4} \int \left(1 + 2 \cos 4x + \frac{1 + \cos 8x}{2}\right) dx = \frac{1}{8} \int (3 + 4 \cos 4x + \cos 8x) dx$$
$$= \frac{1}{8} \left[ 3x + 4 \frac{\sin 4x}{4} + \frac{\sin 8x}{8} \right] + C$$
$$= \mathbf{\frac{3}{8} x + \frac{1}{8} \sin 4x + \frac{1}{64} \sin 8x + C}$$
Part 2: Integration by Simplification and Substitution (12-22)
12. $\int \frac{\sin^2 x}{1 + \cos x} dx$
Use $\sin^2 x = 1 – \cos^2 x$:
$$\int \frac{1 – \cos^2 x}{1 + \cos x} dx = \int \frac{(1 – \cos x)(1 + \cos x)}{1 + \cos x} dx$$
$$= \int (1 – \cos x) dx = \mathbf{x – \sin x + C}$$
13. $\int \frac{\cos 2x – \cos 2\alpha}{\cos x – \cos \alpha} dx$
Use $\cos 2\theta = 2 \cos^2 \theta – 1$:
$$\int \frac{(2 \cos^2 x – 1) – (2 \cos^2 \alpha – 1)}{\cos x – \cos \alpha} dx = \int \frac{2 \cos^2 x – 2 \cos^2 \alpha}{\cos x – \cos \alpha} dx$$
$$= 2 \int \frac{(\cos x – \cos \alpha)(\cos x + \cos \alpha)}{\cos x – \cos \alpha} dx = 2 \int (\cos x + \cos \alpha) dx$$
Since $\alpha$ is a constant, $\cos \alpha$ is a constant.
$$= 2 [\sin x + (\cos \alpha) x] + C = \mathbf{2 \sin x + 2x \cos \alpha + C}$$
14. $\int \frac{\cos x – \sin x}{1 + \sin 2x} dx$
Use $1 + \sin 2x = \cos^2 x + \sin^2 x + 2 \sin x \cos x = (\cos x + \sin x)^2$:
$$\int \frac{\cos x – \sin x}{(\cos x + \sin x)^2} dx$$
Let $u = \cos x + \sin x$.
Then $du = (-\sin x + \cos x) dx$.
$$\int \frac{du}{u^2} = \int u^{-2} du = \frac{u^{-1}}{-1} + C = -\frac{1}{u} + C$$
$$= \mathbf{-\frac{1}{\cos x + \sin x} + C}$$
15. $\int \tan^3 2x \sec 2x dx$
Rewrite $\tan^3 2x = \tan^2 2x \tan 2x = (\sec^2 2x – 1) \tan 2x$.
$$\int (\sec^2 2x – 1) \sec 2x \tan 2x dx = \int (\sec^2 2x \sec 2x \tan 2x – \sec 2x \tan 2x) dx$$
Let $u = \sec 2x$.
Then $du = \sec 2x \tan 2x \cdot 2 dx$, so $\sec 2x \tan 2x dx = \frac{1}{2} du$.
$$\int (u^2 – 1) \frac{1}{2} du = \frac{1}{2} \left[ \frac{u^3}{3} – u \right] + C$$
$$= \mathbf{\frac{1}{6} \sec^3 2x – \frac{1}{2} \sec 2x + C}$$
16. $\int \tan^4 x dx$
Rewrite $\tan^4 x = \tan^2 x \tan^2 x = \tan^2 x (\sec^2 x – 1) = \tan^2 x \sec^2 x – \tan^2 x$.
$$\int (\tan^2 x \sec^2 x – (\sec^2 x – 1)) dx = \int (\tan^2 x \sec^2 x – \sec^2 x + 1) dx$$
- For $\int \tan^2 x \sec^2 x dx$, let $u = \tan x$, $du = \sec^2 x dx$. $\int u^2 du = u^3/3$.$$\frac{\tan^3 x}{3} – \tan x + x + C$$$$= \mathbf{\frac{1}{3} \tan^3 x – \tan x + x + C}$$
17. $\int \frac{\sin^3 x + \cos^3 x}{\sin^2 x \cos^2 x} dx$
Separate the fraction:
$$\int \left( \frac{\sin^3 x}{\sin^2 x \cos^2 x} + \frac{\cos^3 x}{\sin^2 x \cos^2 x} \right) dx$$
$$\int \left( \frac{\sin x}{\cos^2 x} + \frac{\cos x}{\sin^2 x} \right) dx = \int \left( \tan x \sec x + \cot x \csc x \right) dx$$
$$= \mathbf{\sec x – \csc x + C}$$
18. $\int \frac{\cos 2x + 2 \sin^2 x}{\cos^2 x} dx$
Use $\cos 2x = \cos^2 x – \sin^2 x$:
$$\int \frac{(\cos^2 x – \sin^2 x) + 2 \sin^2 x}{\cos^2 x} dx = \int \frac{\cos^2 x + \sin^2 x}{\cos^2 x} dx$$
$$\int \frac{1}{\cos^2 x} dx = \int \sec^2 x dx = \mathbf{\tan x + C}$$
19. $\int \frac{1}{\sin x \cos^3 x} dx$
Rewrite the numerator as $1 = \sin^2 x + \cos^2 x$:
$$\int \frac{\sin^2 x + \cos^2 x}{\sin x \cos^3 x} dx = \int \left( \frac{\sin^2 x}{\sin x \cos^3 x} + \frac{\cos^2 x}{\sin x \cos^3 x} \right) dx$$
$$\int \left( \frac{\sin x}{\cos^3 x} + \frac{1}{\sin x \cos x} \right) dx$$
- Term 1: $\frac{\sin x}{\cos x} \frac{1}{\cos^2 x} = \tan x \sec^2 x$. Let $u = \tan x$.
- Term 2: Multiply $\frac{1}{\sin x \cos x}$ by $\frac{\cos x}{\cos x}$ to get $\frac{\cos x}{\sin x \cos^2 x} = \cot x \sec^2 x$. (Not helpful)
- Alternative for Term 2: Multiply $\frac{1}{\sin x \cos x}$ by $\frac{2}{2}$ to get $\frac{2}{\sin 2x} = 2 \csc 2x$.
$$\int (\tan x \sec^2 x + 2 \csc 2x) dx$$
$$= \frac{\tan^2 x}{2} + 2 \left( -\frac{1}{2} \log|\csc 2x + \cot 2x| \right) + C$$
$$= \mathbf{\frac{1}{2} \tan^2 x – \log|\csc 2x + \cot 2x| + C}$$
(Alternative using $\tan x$ substitution: $\frac{1}{2} \tan^2 x + \log|\tan x| + C$.)
20. $\int \frac{\cos 2x}{(\cos x + \sin x)^2} dx$
Use $\cos 2x = \cos^2 x – \sin^2 x = (\cos x – \sin x)(\cos x + \sin x)$:
$$\int \frac{(\cos x – \sin x)(\cos x + \sin x)}{(\cos x + \sin x)^2} dx = \int \frac{\cos x – \sin x}{\cos x + \sin x} dx$$
Let $u = \cos x + \sin x$.
Then $du = (-\sin x + \cos x) dx$.
$$\int \frac{du}{u} = \log|u| + C = \mathbf{\log|\cos x + \sin x| + C}$$
21. $\int \sin^{-1} (\cos x) dx$
Use the identity $\sin^{-1}(\cos x) = \frac{\pi}{2} – \cos^{-1}(\cos x) = \frac{\pi}{2} – x$, for $x \in [0, \pi]$.
$$\int \left(\frac{\pi}{2} – x\right) dx = \mathbf{\frac{\pi}{2} x – \frac{x^2}{2} + C}$$
22. $\int \frac{1}{\cos(x – a) \cos(x – b)} dx$
This integral requires a special trick using a constant identity in the numerator.
Let $A = x – b$ and $B = x – a$. Then $A – B = a – b$.
Multiply and divide by $\sin(a – b)$:
$$\frac{1}{\sin(a – b)} \int \frac{\sin((x – b) – (x – a))}{\cos(x – a) \cos(x – b)} dx$$
Use $\sin(A – B) = \sin A \cos B – \cos A \sin B$:
$$\frac{1}{\sin(a – b)} \int \frac{\sin(x – b) \cos(x – a) – \cos(x – b) \sin(x – a)}{\cos(x – a) \cos(x – b)} dx$$
Separate the fraction:
$$\frac{1}{\sin(a – b)} \int \left[ \frac{\sin(x – b)}{\cos(x – b)} – \frac{\sin(x – a)}{\cos(x – a)} \right] dx$$
$$\frac{1}{\sin(a – b)} \int [\tan(x – b) – \tan(x – a)] dx$$
$$\frac{1}{\sin(a – b)} [-\log|\cos(x – b)| – (-\log|\cos(x – a)|)] + C$$
$$= \mathbf{\frac{1}{\sin(a – b)} \log \left| \frac{\cos(x – a)}{\cos(x – b)} \right| + C}$$
Part 3: Multiple Choice Questions
23. $\int \frac{\sin^2 x – \cos^2 x}{\sin^2 x \cos^2 x} dx$
Separate the fraction:
$$\int \left( \frac{\sin^2 x}{\sin^2 x \cos^2 x} – \frac{\cos^2 x}{\sin^2 x \cos^2 x} \right) dx = \int \left( \frac{1}{\cos^2 x} – \frac{1}{\sin^2 x} \right) dx$$
$$= \int (\sec^2 x – \csc^2 x) dx = \tan x – (-\cot x) + C$$
$$= \mathbf{\tan x + \cot x + C}$$
The correct answer is (A) $\tan x + \cot x + C$.
24. $\int \frac{e^x (1 + x)}{\cos^2 (x e^x)} dx$
Let $u = x e^x$.
Find $du$ using the product rule: $du = (1 \cdot e^x + x \cdot e^x) dx = e^x (1 + x) dx$.
The integral is $\int \frac{du}{\cos^2 u} = \int \sec^2 u du$.
$$\int \sec^2 u du = \tan u + C = \mathbf{\tan (x e^x) + C}$$
The correct answer is (B) $\tan (x e^x) + C$.