### A solution by Sourav De

Evaluate $\int \frac{d\theta}{(\sqrt{\sin \theta} + \sqrt{\cos \theta})^4}$

$\int \frac{d\theta}{(\sqrt{\sin \theta} + \sqrt{\cos \theta})^4}$

$=\int \frac{d\theta}{(\cos \theta)^2(\sqrt{\tan \theta} + 1)^4}$

$=\int \frac{\sec^2 \theta \; d\theta}{(\sqrt{\tan \theta} + 1)^4}$

Substitute $u^2 = \tan \theta$ and $2u \; du = \sec^2 \theta \; d\theta$

$\int \frac{\sec^2 \theta \; d\theta}{(\sqrt{\tan \theta} + 1)^4}$

$=\int \frac{2u \; du}{(1 + u)^4}$

$=2 \int \left( \frac{1}{(1 + u)^3} - \frac{1}{(1 + u)^4} \right) \; du$

Substitute $z = 1 + u$ and $du = dz$

$=2 \int \left( z^{-3} - z^{-4} \right) \; du$

$= -z^{-2} + \frac{2}{3}z^{-3} + C$

$= \frac{-1}{(1+u)^2} + \frac{2}{3(1 + u)^3} + C$

$= \frac{-1}{(1+\sqrt{\tan \theta})^2} + \frac{2}{3(1 + \sqrt{\tan \theta})^3} + C$