Indefinite integral of $\arctan{\sqrt{1-x^{2}}}$ All is in the title: what is the antiderivative of $x\mapsto \arctan{\sqrt{1-x^{2}}}$ ?
I'm supposed to tutor younger students taking an integration class, and this is one of their exercises. I strongly dislike this kind of math and, consequently, I'm not very good at it, but I'd really like to do a good job, so thanks for your help.
 A: Here's the end of the story:
$$\int\arctan(\cos u)\cdot\cos (u)\,\mathrm d\mkern1mu u=\ \arctan(\cos u)\cdot\sin(u)\ +\ \int\frac{\sin^2u}{1+\cos^2u}\,\mathrm du. $$
Now $\,\displaystyle \sin^2u=\frac{\tan^2u}{1+\tan^2u},\quad \cos^2u=\frac1{1+\tan^2u}$. Set $t=\tan u$. We have $\mathrm d\mkern1mu u=\dfrac{\mathrm d\mkern1mu t}{1+t^2}$ so the integral becomes
\begin{align*}
\int\frac{t^2}{(2+t^2)(1+t^2)}\mathrm d\mkern1mu t&=\int\frac{2\,\mathrm d\mkern1mu t}{(2+t^2)}-\int\frac{\mathrm d\mkern1mu t}{(1+t^2)}\\
&=\sqrt2\,\arctan\frac t{\sqrt2}-\arctan t \\
&=\sqrt2\,\arctan\frac{\tan u}{\sqrt2}- u \quad (+\,\mathrm{constant})
\end{align*}\
A: Well, $\sqrt{1-x^2}$ suggests a substitution like $x=\sin(u)$.
Then $dx=\cos(u)\cdot du$ and
$$\int \arctan(\sqrt{1-x^2})\,dx\ =\ \int\arctan(\cos u)\cdot\cos (u)\,du\,.$$
We would be more happy if one of these terms was $\sin u$, as then we could use the formula for inner function.
So, let's use integration by parts:
$$\int\arctan(\cos u)\cdot\cos (u)\,du\ =\ \arctan(\cos u)\cdot\sin(u)\ -\ \int\frac1{1+\cos^2u}\cdot\sin(u)\,du$$
Now use substition $y=\cos(u)=\sqrt{1-x^2}$ and finally write it all back to $x$.
