Integral $\int_{0}^{\pi/2} \arctan \left(2\tan^2 x\right) \mathrm{d}x$

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?

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 $$ I=\int_0^2I'(a)da=\pi\arctan(1+\sqrt{2a})\Big]_0^2=-\frac{\pi^2}{4}+\pi\arctan 3=\frac{\pi}{2}\arctan\frac{4}{3}.$$

Here is a result avoiding differentiation with respect to a parameter.

Set$$ I(\alpha):= \int_{0}^{\Large\frac{\pi}{2}} \arctan \left(\frac{2\alpha \:\sin^2 x}{\alpha^2-1+\cos^2 x}\right)\: \mathrm{d}x, \quad \alpha>0. $$ Then $$ I(\alpha)= \pi \arctan \left(\frac{1}{2\alpha}\right) \quad ({\star}) $$

With $ \alpha:=1$, we get $$ \int_{0}^{\Large\frac{\pi}{2}} \arctan \left(2 \tan^2 x\right) \mathrm{d}x = \pi \arctan \left( \frac{1}{2} \right). $$ To obtain $({\star})$ use the standard evaluation extended to complex numbers