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Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 10 Maths Textbook Solution.

Answers (1)

Answer:

-\cot x-\sin 2 x-2 x+c

Given:

\int \operatorname{cosec}^{2} x \cos ^{2} 2 x d x

Hint:

Use trigonometric identities and integrate it by parts.

Solution:  

\int \operatorname{cosec}^{2} x \cdot \cos ^{2} 2 x d x

=\int \operatorname{cosec}^{2} x\left(1-2 \sin ^{2} x\right)^{2} d x \quad\left[\because \cos 2 \mathrm{x}=1-2 \sin ^{2} \mathrm{x}\right]

=\int \operatorname{cosec}^{2} x\left(1+4 \sin ^{4} x-4 \sin ^{2} x\right) d x \quad\left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right]

=\int\left(\operatorname{cosec}^{2} x+4 \sin ^{2} x-4\right) d x \quad\left[\because \sin x=\frac{1}{\operatorname{cosec} x}\right]

=\int \operatorname{cosec}^{2} x d x+2 \int 1-\cos 2 x d x-4 \int d x

=-\cot x+2 x-\sin 2 x-4 x+c

=-\cot x-\sin 2 x-2 x+c 

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