<s>[INST] 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.
 [/INST] HINT:
Metheod$\#1:$
Let $\sqrt{\dfrac{1-x^2}{1+x^2}}=y\implies \dfrac{1-x^2}{1+x^2}=y^2\implies  x^2=\dfrac{1-y^2}{1+y^2}=\dfrac{2-(1+y^2)}{1+y^2}=\cdots$
$$2x\ dx=-2\dfrac{2y\ dy}{(1+y^2)^2}$$
$$\dfrac{d(y/1+y^2)}{dy}=?$$
Metheod$\#2:$
Let $x^2=\cos2y\implies0\le2y\le\pi$ and $\sqrt{\dfrac{1-x^2}{1+x^2}}=+\tan y$
$x\ dx=-2\sin2y\ dy=-4\sin y\cos y\ dy$ 
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