Help with the integral $\int x\sqrt{\frac{1-x^2}{1+x^2}}dx$ I would like to know what is $$\int x\sqrt{\frac{1-x^2}{1+x^2}}dx.$$  I put $x=\tan(y)$ to get integral of $\displaystyle \int \frac{\sin(y)}{\cos^3(y)}.\sqrt{\cos(2y)}dy$ I don't know whether $\sin(x)=t$ is a good substitution.

In the same spirit as previous answer considering $$I = \int x\, \sqrt{\frac{1-x^2}{1+x^2}}\, dx$$ What I prefer is to get rid of the radical as early as possible.
So, let $$\sqrt{\frac{1-x^2}{1+x^2}}=u \implies x=\frac{\sqrt{1-u^2}}{\sqrt{u^2+1}}\implies dx=-\frac{2 u}{\sqrt{1-u^2} \left(u^2+1\right)^{3/2}}$$ This makes $$I=-\int\frac{2 u^2}{\left(u^2+1\right)^2}\,du$$ Now, $$\frac{ u^2}{\left(u^2+1\right)^2}=\frac{ u^2+1-1}{\left(u^2+1\right)^2}=\frac{1}{u^2+1}-\frac{1}{\left(u^2+1\right)^2}$$
