Question

Evaluate : $\int x\cdot \sqrt{\frac{1-x^2}{1+x^2}}dx.$

Answer 1

$\begin{array}{c}GivenI={\int }^{}x\sqrt{\frac{1-x^2}{1+x^2}}dx\\ Let{x}^2=t\\ 2xdx=dt\\ xdx=\frac{dt}{2}\\ ThenI={\int }^{}\sqrt{\frac{1-t}{1+t}}\frac{dt}{2}\\ =\frac{1}{2}{\int }^{}\frac{1-t}{\sqrt{1-t^2}}dt\\ =\frac{1}{2}{\int }^{}\frac{dt}{\sqrt{1-t^2}}-\frac{1}{2}{\int }^{}\frac{tdt}{\sqrt{1-t^2}}\\ Lett=\sin \theta \\ dt=\cos \theta d\theta \\ =\frac{1}{2}{\int }^{}\frac{\cos d\theta }{\cos \theta }-\frac{1}{2}{\int }^{}\frac{\sin \theta \cos \theta }{\cos \theta }d\theta \\ =\frac{1}{2}{\int }^{}d\theta -\frac{1}{2}{\int }^{}\cos \theta d\theta \\ =\frac{1}{2}\theta -\frac{1}{2}\sin \theta +C\\ =\frac{1}{2}\left({\sin }^{-1}t-t\right)+C\\ =\frac{1}{2}({\sin }^{-1}{x}^2-x^2)+C\end{array}$