Question

Show that $\int x\sqrt{\left(\frac{1-x^2}{1+x^2}\right)}dx=[{\sin }^{-1}{x}^2+\sqrt{1-x^4}].$

Answer 1

Let $I=\int x\sqrt{\frac{1-x^2}{1+x^2}}dx$
Put $t=x^2\Rightarrow dt=2xdx$
$I=\frac{1}{2}\int \sqrt{\frac{1-t}{t+1}}dt$
Put $s=\frac{1-t}{t+1}\Rightarrow ds=(-\frac{1-t}{{(t+1)}^2}-\frac{1}{t+1})dt$
$I=-\int \frac{\sqrt{s}}{{(-s-1)}^2}ds$
Put $p=\sqrt{s}\Rightarrow dp=\frac{1}{2\sqrt{s}}dx$
$I=-2\int \frac{p^2}{{(-p^2-1)}^2}dp=-2\int (\frac{1}{p^2+1}-\frac{1}{{(p^2+1)}^2})dp=-{\tan }^{-1}p+\sin \left({\tan }^{-1}p\right)\cos \left({\tan }^{-1}p\right)$
Therefore $I=\frac{1}{2}(\sqrt{1-x^2}\sqrt{1+x^2}-{\tan }^{-1}\sqrt{\frac{1-x^2}{1+x^2}})$