$\int \{\sqrt{\tan x}+\sqrt{\cot x}\}dx$ is equal to 

Solution:

We have,

$\begin{array}{l}I=\int \{\sqrt{\tan x}+\sqrt{\cot x}\}dx=\int \frac{\sin x+\cos x}{\sqrt{\sin x\cos x}}dx\\ \Rightarrow I=\sqrt{2}\int \frac{\sin x+\cos x}{\sqrt{2\sin x\cos x}}dx\\ \Rightarrow I=\sqrt{2}\int \frac{1}{\sqrt{1-(\sin x-\cos x{)}^2}}d(-\cos x+\sin x)\\ \Rightarrow I=\sqrt{2}{\sin }^{-1}(\sin x-\cos x)+C.\end{array}$