Question:

Let $I = \int_{0}^{\infty} (\cot^{-1} x)^2 dx \Rightarrow$ Let $x = \cot \theta \Rightarrow dx = -\csc^2 \theta d\theta$<br/>$\therefore I = \int_{\pi/2}^{0} \theta^2 (-\csc^2 \theta) d\theta \Rightarrow I = \int_{0}^{\pi/2} \theta^2 (\csc^2 \theta) d\theta$<br/>$= (\theta^2 (-\cot \theta))_0^{\pi/2} + 2 \int_{0}^{\pi/2} \theta \cot \theta d\theta \Rightarrow I = 0 + 2 \int_{0}^{\pi/2} \theta \cot \theta d\theta$<br/>$= (2\theta \ln \sin \theta)_0^{\pi/2} - 2 \int_{0}^{\pi/2} \ln \sin \theta d\theta$<br/>Standard result: $\int_{0}^{\pi/2} \ln \sin \theta d\theta = -\frac{\pi}{2} \ln 2$<br/>$= 0 - 2 \times \left( -\frac{\pi}{2} \right) \ln 2 = \pi \ln 2$.