Evaluate $\int x \sqrt{\frac{1-x}{1+x}} d x$
Let, $x=\sin t$
Differentiate both side with respect to t
$\frac{d x}{d t}=\cos t \Rightarrow d x=\cos t \mathrm{dt}$
$y=\int \sin t \sqrt{\frac{1-\sin t}{1+\sin t}} \cos t d t$
$y=\int \sin t \sqrt{\frac{(1-\sin t)(1-\sin t)}{(1+\sin t)(1-\sin t)}} \cos t d t$
$y=\int \sin t(1-\sin t) d t$
$y=\int \sin t d t-\int \sin ^{2} t d t$
$y=-\cos t-\int \frac{1-\cos 2 t}{2} d t$
$y=-\cos t-\left(\frac{t}{2}-\frac{\sin 2 t}{4}\right)+c$
Again put $t=\sin ^{-1} x$
$y=-\cos \left(\sin ^{-1} x\right)-\left(\frac{\left(\sin ^{-1} x\right)}{2}-\frac{\sin 2\left(\sin ^{-1} x\right)}{4}\right)+c$
$y=-\sqrt{1-x^{2}}-\frac{\sin ^{-1} x}{2}+\frac{x \sqrt{1-x^{2}}}{2}+c$
$y=\left(\frac{x}{2}-1\right) \sqrt{1-x^{2}}-\frac{1}{2} \sin ^{-1} x+c$