Question

Evaluate $\int \frac{x^{2}dx}{\sqrt{1-x^2}}$

Answer 1

Let $x=\sin \theta$

$\theta ={\sin }^{-1}xdx=\cos \theta d\theta or\int \frac{\sin {}^2\theta }{\sqrt{1-\sin {}^2\theta }}\times cos\theta d\theta or\int \frac{\sin {}^2\theta }{\sqrt{\cos {}^2}\theta }\cos \theta d\theta or\int \frac{\sin {}^2\theta }{\cos \theta }\times \cos \theta d\theta$ by using the formula $[\frac{1-\cos 2\theta }{2}=\sin {}^2\theta ]or\int \frac{1-\cos 2\theta }{2}d\theta =\int \frac{1}{2}d\theta -\frac{1}{2}\int \cos 2\theta d\theta =\frac{\theta }{2}-\frac{1}{4}\sin 2\theta +C=\frac{\theta }{2}-\frac{1}{4}\times 2\sin \theta \cos \theta +C=\frac{1}{2}[{\sin }^{-1}x-\sin \left({\sin }^{-1}x\right)\cos \left({\sin }^{-1}x\right)]+C=\frac{1}{2}\left[{\sin }^{-1}x-x\sqrt{1-x^2}\right]+C$