Question

Evaluate:

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

Answer 1

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

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

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

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

Put $\sin x-\cos x=t$

$(\cos x+\sin x)=\frac{dt}{dx}$

Also, $(\sin x-\cos x{)}^2=t^2$

${\sin }^{2}x+{\cos }^{2}x-2\sin x\cos x=t^2$

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

$\sin x\cos x=\frac{1-t^2}{2}$

Therefore,

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

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

$I=\sqrt{2}{\sin }^{-1}\left(t\right)+C$

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