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I've just begun to learn integration which makes me a little nervous! Here's a question I'm having a problem with.

Also my first time trying to use LaTeX. I apologise for any discrepancies.

Compute: $$ \int \frac{{x}~{\cos^{-1}(x)}}{\sqrt{1-{x^2}}}~\mathrm{d}x $$

Here's what I did:

Substitute $ u = \cos^{-1}(x) $. So, $ -~\mathrm{d}u = \frac{1}{\sqrt{1-x^2}}~\mathrm{d}x$. Also, I think it's also correct (please correct me if not) that $ x = \cos(u) $. This could be my mistake.

$$ = -\int u~\cos(u)~\mathrm{d}u $$

Using integration by parts for $ f(x) = u $, $ f'(x) = 1 $, $ g'(x) = \cos(u) $ and $ g(x) = \sin(u) $,

$$ = -~(u~\sin(u) - \int \sin(u)~\mathrm{d}u) $$ $$ = -~u~\sin(u) - \cos(u) + C $$

Substituing back,

$$ -~\cos^{-1}(x)~\sin(\cos^{-1}(x)) - \cos(\cos^{-1}(x)) + C $$

I understand integration by parts could've been applied directly in the very beginning. But my first instinct when I solved was this. Is it in any way incorrect? I appreciate your time.

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  • $\begingroup$ write \cos for $\cos$ $\endgroup$ – Frank Vel Nov 10 '14 at 17:59
  • $\begingroup$ Thanks for pointing that out @fvel. $\endgroup$ – Shreyas Nov 10 '14 at 18:02
  • $\begingroup$ are you substituting $u$ in your $\sin(u)$ at the last step? $\endgroup$ – Frank Vel Nov 10 '14 at 18:03
  • $\begingroup$ Ah, thanks for pointing that out too @fvel. Yes, in my work. Forgot to do that here! $\endgroup$ – Shreyas Nov 10 '14 at 18:05
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    $\begingroup$ @Fly by Night A few days ago, yes. There's too much good content online! I still need tremendous amounts of practice though. Thanks for the encouragement! $\endgroup$ – Shreyas Nov 10 '14 at 18:17
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Great work! The problem is just when you substitute back to $x$. Recall that $u = \cos^{-1}x$ and $x = \cos u$, so: $$ \sin u = \sqrt{\sin^2 u} = \sqrt{1 - \cos^2 u} = \sqrt{1 - x^2} $$ Hence, we obtain: $$ -u \sin u - \cos u + C = -(\cos^{-1}x)\sqrt{1 - x^2} - x + C $$

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  • $\begingroup$ The beauty of it is quite amazing isn't it! Thank you. $\endgroup$ – Shreyas Nov 10 '14 at 18:12

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