TITLE: Integral $\int_{0}^{\pi/2} \arctan \left(2\tan^2 x\right) \mathrm{d}x$
QUESTION [15 upvotes]: The following integral may seem easy to evaluate ...

$$
\int_{0}^{\Large\frac{\pi}{2}} \arctan \left(2 \tan^2 x\right) \mathrm{d}x = \pi \arctan \left( \frac{1}{2} \right).
$$

Could you prove it?

REPLY [4 votes]: My answer is different from that you gave. Let
$$ I(a)=\int_0^{\frac{\pi}{2}}\arctan(a\tan^2x)dx. $$
Than $I(0)=0$ and
\begin{eqnarray}
I'(a)&=&\int_0^{\frac{\pi}{2}}\frac{\tan^2x}{1+a^2\tan^4x}dx\\
&=&\int_0^\infty\frac{u^2}{(1+u^2)(1+a^2u^4)}du\\
&=&\frac{1}{1+a^2}\int_0^\infty\left(-\frac{1}{1+u^2}+\frac{1+a^2u^2}{1+a^2u^4}\right)du\\
&=&\frac{1}{1+a^2}\left(-\int_0^\infty\frac{1}{1+u^2}du+\int_0^\infty\frac{1}{1+a^2u^4}du+\int_0^\infty\frac{a^2u^2}{1+a^2u^4}\right)du\\
&=&\frac{1}{1+a^2}\left(-\frac{\pi}{2}+\frac{\pi}{2\sqrt{2}\sqrt{a}}+\frac{\sqrt{a}\pi}{2\sqrt{2}}\right)
\end{eqnarray}
and hence
$$ I=\int_0^2I'(a)da=\pi\arctan(1+\sqrt{2a})\Big]_0^2=-\frac{\pi^2}{4}+\pi\arctan 3=\frac{\pi}{2}\arctan\frac{4}{3}.$$