Evaluating $\int x\sin^{-1}x dx$

Question

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:

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

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

Please help!!!

Answer 1

Let $x=\cos{t}$, where $t\in[0,\pi]$.

Hence, $\sqrt{1-x^2}dx=-\sin^{2}tdt=-\frac{1-\cos2t}{2}dt$

Answer 2

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, $y = \sin^{-1}x$.

Answer 3

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 $$I:=x\sqrt{1-x^2}-\int\frac{dx}{\sqrt{1-x^2}}+\int\sqrt{1-x^2}\,dx=x\sqrt{1-x^2}-\arcsin x-I.$$

Answer 4

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!