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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 9 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 9 Maths Textbook Solution.

Answers (1)

Answer:

\sin 2 x+\tan x-2 x+c

Given:

\int \sec ^{2} x \cos ^{2} 2 x d x

Hint:

Use trigonometry identity.

Solution:   

\int \sec ^{2} x \cos ^{2}(2 x) d x

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

     =\int \sec ^{2} x\left(4 \cos ^{4} x-4 \cos ^{2} x+1\right) d x

  =\int 4 \cos ^{2} x-4+\sec ^{2} x d x \quad\left[\because \frac{1}{\sec ^{2} x}=\cos ^{2} x\right]

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

=2 \int 2 \cos ^{2} x d x-4 \int d x+\int \sec ^{2} x d x

=2 \int(1+\cos 2 \mathrm{x}) \mathrm{dx}-4 \int \mathrm{dx}+\int \sec ^{2} \mathrm{x} \mathrm{dx}

=2 x+2 \frac{\sin 2 x}{2}-4 x+\tan x+c

=\sin 2 x-2 x+\tan x+c

 

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