$$\begin {aligned} f(a)&:=\int _{0}^{\frac {\pi }2}\arctan \left (a\tan ^{2}x\right )dx\\ &=\int _{0}^{\infty }\frac {\arctan ax^{2}}{1+x^{2}}dx\\ f'(a)&=\int _{0}^{\infty }\frac {x^{2}}{\left (1+x^{2}\right )\left (1+a^{2}x^{4}\right )}dx\\ &=\frac {1}{1+a^{2}}\int _{0}^{\infty }\left (\frac {1+a^{2}x^{2}}{1+a^2x^{4}}-\frac {1}{1+x^{2}}\right )dx\\ &=\frac {1}{1+a^{2}}\left (\frac {\pi }{2\sqrt {2a}}+\frac {\pi \sqrt a}{2\sqrt 2}-\frac {\pi }2\right )\\ &=\frac {\pi }2\frac {1}{1+a^{2}}\left (-1+\frac {1}{\sqrt {2a}}+\frac {\sqrt 2}{\sqrt a}\right )\\ f(2)&=\frac {\pi }2\int _{0}^{2}\frac {1}{1+t^{2}}\left (-1+\frac {1}{\sqrt {2t}}+\frac {\sqrt t}{\sqrt 2}\right )dt\\ &=\frac {\pi }2\left (-\arctan 2+\frac {1}{\sqrt 2}\int _{0}^{2}\frac {\sqrt t+\frac {1}{\sqrt t}}{1+t^{2}}dt\right )\\ &=\frac {\pi }2\left (-\arctan 2+\sqrt 2\int _{0}^{\sqrt 2}\frac {1+t^{2}}{1+t^{4}}dt\right )\\ &=\frac {\pi }2\left (-\arctan 2+\sqrt 2\Re \int _{0}^{\sqrt 2}\frac {1-i}{t^{2}-i}dt\right )\\ &=\frac {\pi }2\left (-\arctan 2 +\sqrt 2\Re \left [\frac {1-i}{2\sqrt i}\int _{0}^{\sqrt 2}\left (\frac {1}{t-\sqrt i}-\frac {1}{t+\sqrt i}\right )dt\right ]\right )\\ &=\frac {\pi }2\left (-\arctan 2+\Re \left [\frac {1}{i}\ln \frac {t-\sqrt i}{t+\sqrt i}\right ]_0^{\sqrt 2}\right )\\ &=\frac {\pi }2\left (-\arctan 2+\Im \ln \frac {\sqrt 2-\sqrt i}{\sqrt 2+\sqrt i}-\Im \ln \frac {0-\sqrt i}{0+\sqrt i}\right )\\ &=\frac {\pi }2\left (-\arctan 2+\Im \ln \left (\frac {1-2i}5\right )+\pi \right )\\ &=\frac {\pi }2\left (-\arctan 2-\arctan 2+\pi \right )\\ &=\pi \left (\frac {\pi }2-\arctan 2\right )\\ &=\textcolor {blue}{\pi \arctan \frac {1}2}. \end {aligned}$$