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Please Solve R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise  Revision Exercise Question 94 Maths Textbbok Solution.

Answers (1)

Answer:

\frac{x \tan 2 x}{2}-\frac{1}{4} \log |\sec 2 x|+c

Hint:

You must know about integration of sec x & tan x.

Given:

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

Solution:

\int x \sec ^{2} 2 x d x \quad\left(\int1.11 d x=I \int I 1 d x-\frac{d}{d x} 1 . \int I I d x\right)

   x\left(\frac{\tan 2 x}{2}\right)-\frac{1}{2} \int \tan 2 x d x

\frac{x \tan 2 x}{2}-\frac{1}{2} \int \tan 2 x d x \quad\left(\int \tan 2 x=\log \left|\frac{\sec 2 x}{2}\right|\right)

\frac{x \tan 2 x}{2}-\frac{1}{2} \log \left|\frac{\sec 2 x}{2}\right|+c

\frac{x \tan 2 x}{2}-\frac{1}{4} \log |\sec 2 x|+c

 

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