# Computing $\int \frac{{x}~{\cos^{-1}(x)}}{\sqrt{1-{x^2}}}~\mathrm{d}x$.

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.

• write \cos for $\cos$ – Frank Vel Nov 10 '14 at 17:59
• Thanks for pointing that out @fvel. – Shreyas Nov 10 '14 at 18:02
• are you substituting $u$ in your $\sin(u)$ at the last step? – Frank Vel Nov 10 '14 at 18:03
• Ah, thanks for pointing that out too @fvel. Yes, in my work. Forgot to do that here! – Shreyas Nov 10 '14 at 18:05
• @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! – Shreyas Nov 10 '14 at 18:17

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: $$-u \sin u - \cos u + C = -(\cos^{-1}x)\sqrt{1 - x^2} - x + C$$