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Provide Solution For R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 117 Maths Textbook Solution.

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Answer: $\left(\sin ^{-1} x\right) \cdot \frac{1}{\sqrt{1-x^{2}}}-\frac{1}{2} \log \frac{1+x}{1-x}+C$

Hint: to solve this equation, we have to assume x as $\sin \theta$ and then do byparts method

Given: $\int \frac{x \sin ^{-1} x}{\left(1-x^{2}\right)^{3 / 2}} d x$

Soluton: Let $x=\sin \theta$

$d x=\cos \theta d \theta$

$I=\int \frac{\sin \theta(\theta \cos \theta d \theta)}{\left(1-\sin ^{2} \theta\right)^{\frac{3}{2}}}$

$\int \frac{(\sin \theta)(\theta) \cos \theta d \theta}{\cos ^{3} \theta}$

$\int \frac{\sin \theta}{\cos ^{2} \theta} \theta d \theta$

$I=\int \theta \cdot \sec \theta \tan \theta d \theta$

$\left[\int \sec \theta \tan \theta d \theta=\int d \sec \theta=\sec \theta\right]$

$I=\theta \sec \theta \int 1 \cdot \sec \theta d \theta$

$I=\theta \sec \theta-\ln |\sec \theta+\tan \theta|+C$

$I=\left(\sin ^{-1} x\right) \cdot \frac{1}{\sqrt{1-x^{2}}}-\frac{1}{2} \log \frac{1+x}{1-x}+C$

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