Question

Evaluate :
$\int [\sqrt{tanx}+\sqrt{cotx}]dx$.

Answer 1

$I=\int (\sqrt{\tan x}+\sqrt{\cot x})dx$

$\therefore I=\int \sqrt{\frac{\sin x}{\cos x}}+\sqrt{\frac{\cos x}{\sin x}}dx$

$\therefore I=\int \frac{\sin x+\cos x}{\sqrt{\sin x\cos x}}dx$

$\sin x-\cos x=t$

$\therefore (\cos x+\sin x)dx=dt$

$\therefore 1-2\sin x\cos x=t^2$

$\therefore I=\int \frac{\sqrt{2}}{\sqrt{1-t^2}}dt$

$\therefore I=\sqrt{2}{\sin }^{-1}t+c$

$\therefore I=\sqrt{2}{\sin }^{-1}(\sin x-\cos x)+c$