I solved an ellipse-related problem with a polar coordinate transformation and wondered if the following calculation is possible. $$\begin{align}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{a^2b^2}{b^2\cos^2\theta+a^2\sin^2\theta}d\theta &=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{a^2b^2\sec^2\theta}{b^2+a^2\tan^2\theta}d\theta \ \ (tan\theta \longmapsto t)\\ & =\frac{a^2b^2}{2}\int_1^\infty \frac{dt}{b^2+a^2t^2}\\ & = \frac{a^2b}{4}\int_1^\infty \frac{1}{b+ati}+\frac{1}{b-ati}\ dt =\frac{ab}{4i}\log{\frac{b+ati}{b-ati}}\left.\right|_{t=1}^{t\to\infty}\\ & =\frac{ab}{4i}\cdot i\tan^{-1}\frac{2ab}{a^2-b^2} \\ & =\frac{ab}{4}\tan^{-1}\frac{2ab}{a^2-b^2}\end{align}$$
During the process, $$\log \frac{b+ati}{b-ati} $$ can be calculated because $$\frac{b+ati}{b-ati}=\frac{b^2-a^2t^2+2abi}{b^2+a^2t^2}=e^{i\tan^{-1} \frac{2abt}{b^2-a^2t^2}}$$