A calculus problem by Sriram Radhakrishnan

The integral can be solved using the integration trick (reflection) : $\displaystyle \int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx$ .

$\begin{aligned} I & = \int_0^\frac \pi 2 \log \left(\tan^2 x \right) dx \\ & = \frac 12 \int_0^\frac \pi 2 \log \left(\tan^2 x \right) + \log \left(\tan^2 \left(\frac \pi 2 - x \right) \right) dx \\ & = \frac 12 \int_0^\frac \pi 2 \log \left(\tan^2 x \right) + \log \left(\cot^2 x \right) dx \\ & = \frac 12 \int_0^\frac \pi 2 \log \left(\tan^2 x \right) + \log \left(\frac 1{\tan^2 x} \right) dx \\ & = \frac 12 \int_0^\frac \pi 2 \log \left(\tan^2 x \right) - \log \left(\tan^2 x \right) dx \\ & = \boxed{0} \end{aligned}$