check my solution to indefinite integral problem with arccos

Question

So we had homework it asked us to find $$\int\arccos(x)dx$$

I have found that $$\int\arccos(x)dx=x\arccos (x)+\sqrt{1-x^2}+c$$

Is this right?

Answer 1

Integrate by parts: $$\int \arccos xdx= x\arccos x+\int\frac{x}{\sqrt{1-x^2}}dx=x\arccos x-\sqrt{1-x^2}+C$$

Answer 2

It is approximately right since it should be:

$$\int \arccos(x)\,\mathrm dx=x\arccos(x)\color{red}{\mathbf{-}}\sqrt{1-x^2}+\text{const.}$$

Be careful next time when dealing with minuses and pluses. ;-)

I hope this helps.
Best wishes, $\mathcal H$akim.