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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!!!

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  • $\begingroup$ Using one of the magic words of integration: "Let $x=\sin u$..." $\endgroup$ – J. M. is a poor mathematician Aug 14 '17 at 7:43
  • $\begingroup$ 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 $\endgroup$ – Atul Mishra Aug 14 '17 at 7:46
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    $\begingroup$ How do you differentiate the sine? $\endgroup$ – J. M. is a poor mathematician Aug 14 '17 at 7:47
  • $\begingroup$ Oops I got it ... $\endgroup$ – Atul Mishra Aug 14 '17 at 7:48
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Let $x=\cos{t}$, where $t\in[0,\pi]$.

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

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  • $\begingroup$ Also works, but you now have a dangling minus sign to carry around. $\endgroup$ – J. M. is a poor mathematician Aug 14 '17 at 7:53
  • $\begingroup$ @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. $\endgroup$ – Michael Rozenberg Aug 14 '17 at 7:55
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    $\begingroup$ Well, since this isn't definite integration, we can be slightly sloppy here... $\endgroup$ – J. M. is a poor mathematician Aug 14 '17 at 8:00
  • $\begingroup$ @J. M. isn't a mathematician Without previous condition we must write $\sqrt{\cos^2x}=|\cos{x}|$. $\endgroup$ – Michael Rozenberg Aug 14 '17 at 8:08
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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 $$I:=x\sqrt{1-x^2}-\int\frac{dx}{\sqrt{1-x^2}}+\int\sqrt{1-x^2}\,dx=x\sqrt{1-x^2}-\arcsin x-I.$$

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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!

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  • $\begingroup$ A further simplification would be using $\sin t\cos t=\frac{\sin 2t}{2}$. $\endgroup$ – J. M. is a poor mathematician Aug 14 '17 at 8:09
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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, $y = \sin^{-1}x$.

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

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