Evaluating $\int x\sin^{-1}x dx$ I was integrating $$\int x\sin^{-1}x dx.$$
After applying integration by parts and some rearrangement  I  got stuck at $$\int \sqrt{1-x^2}dx.$$
Now I have two questions:


*

*Please, suggest any further approach from where I have stuck;


*Please, provide an alternative way to solve the original question.

Please help!!!
 A: Let $x=\cos{t}$, where $t\in[0,\pi]$.
Hence, $\sqrt{1-x^2}dx=-\sin^{2}tdt=-\frac{1-\cos2t}{2}dt$
A: By parts,
$$I:=\int\sqrt{1-x^2}\,dx=x\sqrt{1-x^2}+\int\frac{x^2}{\sqrt{1-x^2}}dx.$$
But $x^2=1-(1-x^2)$ and you get
$$I:=x\sqrt{1-x^2}-\int\frac{dx}{\sqrt{1-x^2}}+\int\sqrt{1-x^2}\,dx=x\sqrt{1-x^2}-\arcsin x-I.$$
A: Alternative way: Let $x=\sin t$ so 
$$\int x\sin^{-1}x dx=\int t\sin t\cos t dt$$
by parts $t=u$ and $\sin t\cos t dt=dv$ and finish it!
A: I'll proceed from the point you got stuck, 
$$\int \sqrt{1-x^2} dx$$
First, let $x = \sin y$, and thus $dx = \cos y\,dx$. Now substituting in the above integral, 
$$\int \sqrt{1-x^2} = \int \sqrt{1-\sin^2 y} \, \cos y \, dy  $$
$$= \int \sqrt{\cos^2 y}  \, \cos y \, dy   = \int \cos y \, \cdot \cos y\, dy  $$
$$ = \int \cos^2 y \, dy $$
Now using Pythagorean Identity we can write, 
$$ \cos^2 y = \frac{1}{2} + \frac{\cos 2y}{2} $$
Finally the integration becomes, 
$$\int \frac{1}{2} dy + \int \frac{\cos 2y}{2}dy  $$
Go ahead now. Remember, $y = \sin^{-1}x$.
