Impossible integral? $ \int \sqrt{1-x^2} \arccos ( \sqrt{1-x^2} ) dx $ Is it possible to solve this integral?
$$ \int  \sqrt{1-x^2} \arccos ( \sqrt{1-x^2} ) dx $$
 A: Note $ \arccos ( \sqrt{1-x^2} )=\arcsin x$. 
Then, with $t = \arcsin x$,
$$ \int  \sqrt{1-x^2} \arccos ( \sqrt{1-x^2} ) dx =\int  \sqrt{1-x^2} \arcsin x\> dx$$
$$=\int  t\cos^2t dt  =\frac12\int  t(1+\cos 2t) dt 
= \frac14t^2 +\frac14\int t d(\sin 2t) $$
$$= \frac14t^2 + \frac14t\sin2t+\frac18\cos2t+C$$
A: Take $\cos\theta = \sqrt{1-x^2}$.  This yields $-\sin\theta d\theta = -\frac{x}{\sqrt{1-x^2}}dx = -\frac{\sin\theta}{\cos\theta}dx$ so therefore $dx = \cos\theta d\theta$.  The integral transforms to 
\begin{eqnarray*}
\int \sqrt{1-x^2}\arccos\left (\sqrt{1-x^2}\right )dx & = & \int \cos(\theta)\cdot \theta \cdot \cos\theta d\theta \\
& = & \int \theta \cos^2\theta d\theta \\
& = & \int \frac{\theta}{2}(1+\cos(2\theta)) d\theta \\
& = & \frac{1}{4}\theta^2 + \frac{1}{4}\theta\sin(2\theta) - \frac{1}{4}\int \sin(2\theta)d\theta \\
& = & \frac{1}{4}\theta^2 + \frac{1}{4}\theta\sin(2\theta) + \frac{1}{8}\cos(2\theta) + C \\
& = & \frac{1}{4}\theta^2 + \frac{1}{2}\sin\theta\cos\theta + \frac{1}{8}\cos^2\theta - \frac{1}{8}\sin^2\theta + C \\
& = & \frac{1}{4}\left (\arccos\sqrt{1-x^2}\right )^2 + \frac{1}{2}x\sqrt{1-x^2} -\frac{1}{4}x^2 + \frac{1}{8} +C.
\end{eqnarray*}
