# 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:

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

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

• Using one of the magic words of integration: "Let $x=\sin u$..." – J. M. is a poor mathematician Aug 14 '17 at 7:43
• If I am not mistaking, wouldn't it will become $\cos u~d\sin u$ ? How will I integrate it with respect to $\sin u$?? @ J.M.isn'tamathematician – Atul Mishra Aug 14 '17 at 7:46
• How do you differentiate the sine? – J. M. is a poor mathematician Aug 14 '17 at 7:47
• Oops I got it ... – Atul Mishra Aug 14 '17 at 7:48

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

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

• Also works, but you now have a dangling minus sign to carry around. – J. M. is a poor mathematician Aug 14 '17 at 7:53
• @J. M. isn't a mathematician I think to say $x=\sin{t}$ without $t\in[-\frac{\pi}{2},\frac{\pi}{2}]$ it's very bad. – Michael Rozenberg Aug 14 '17 at 7:55
• Well, since this isn't definite integration, we can be slightly sloppy here... – J. M. is a poor mathematician Aug 14 '17 at 8:00
• @J. M. isn't a mathematician Without previous condition we must write $\sqrt{\cos^2x}=|\cos{x}|$. – Michael Rozenberg Aug 14 '17 at 8:08

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

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 further simplification would be using $\sin t\cos t=\frac{\sin 2t}{2}$. – J. M. is a poor mathematician Aug 14 '17 at 8:09

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

• I think $\sqrt{\cos^2x}=|\cos{x}|$. See please, my solution. – Michael Rozenberg Aug 14 '17 at 8:11
• @MichaelRozenberg Yes absolutely but as it's an indefinite integral we can get irresponsible a bit. – Ankit Panda Aug 14 '17 at 8:14
• I understand you, but I don'd like this thing. – Michael Rozenberg Aug 14 '17 at 8:18