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Provide Solution For R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise 18.25 Question 33 Maths Textbook Solution.

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Answer: =\tan \: x+log\left ( \cos x \right )-\frac{x^{2}}{2}+c

Given: \int x\left ( \frac{\sec 2x-1}{\sec 2x+1} \right )dx

Hint: \int uvdx=u\int vdx-\int \left ( \frac{d}{dx}u\int vdx \right )dx


            \begin{aligned} &I=\int x\left(\frac{1-\cos 2 x}{1+\cos 2 x}\right) d x \\ &I=\int x\left(\frac{2 \sin ^{2} x}{2 \cos ^{2} x}\right) d x \\ &I=\int x\left(\sec ^{2} x-1\right) d x \\ &I=\int x \sec ^{2} x-\int x d x \\ &I=x \int \sec ^{2} x-\int\left[\frac{d}{d x} x \int \sec ^{2} x d x\right] d x-\frac{x^{2}}{2}+c \\ &I=x \tan x-\int \tan x d x-\frac{x^{2}}{2}+c \\ &=x \tan x+\log (\cos x)-\frac{x^{2}}{2}+c \end{aligned}

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