Question

Solve:$\int \sqrt{\tan x}+\sqrt{\cot x}.dx$

Answer 1

Given,

$\int (\sqrt{\tan x}+\sqrt{\cot x})dx=\int (\sqrt{\frac{\sin x}{\cos x}}+\sqrt{\frac{\cos x}{\sin x}})dx=\int \left(\frac{\sin x+\cos x}{\sqrt{2\sin x\cos x}}\right)dx=\int \frac{\sqrt{2}(\sin x+\cos x)}{\sqrt{2\sin x\cos x}}dx=\sqrt{2}\int \frac{\sin x+\cos x}{\sqrt{1-(1-\sin 2x)}}dx=\sqrt{2}\int \frac{\sin x+\cos x}{\sqrt{1-{(\sin x-\cos x)}^2}}dx$

Let, $u=\sin x-\cos x\Rightarrow du=(\cos x+\sin x)dx$

Substituting the values of $u$ and $du$ we get,

$=\sqrt{2}\int \frac{du}{\sqrt{1-u^2}}=\sqrt{2}{\sin }^{-1}u+C=\sqrt{2}{\sin }^{-1}(\sin x-\cos x)+C$

$\therefore \int (\sqrt{\tan x}+\sqrt{\cot x})dx=\sqrt{2}{\sin }^{-1}(\sin x-\cos x)+C.$