Question

If $I=\int (\sqrt{\tan x}+\sqrt{\cot x})dx=f\left(x\right)+c$

• A. $\sqrt{2}{\sin }^{-1}\times (\sin x-\cos x)$
• B. $\sqrt{2}{\sin }^1\times (\sin x-\cos x)$
• C. $\sqrt{2}{\sin }^{1-}\times (\sin x+\cos x)$
• D. $noneofthese$

Answer 1

The correct option is A $\sqrt{2}{\sin }^{-1}\times (\sin x-\cos x)$

Given,

$\int \sqrt{\tan x}+\sqrt{\cot x}dx$

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

put $\sin x-\cos x=u\to du=\sin x+\cos xdx$

$u^2={\sin }^{2}x+{\cos }^{2}x-2\sin x\cos x\Rightarrow \sin x\cos x=\frac{1-u^2}{2}$

$I=\int \frac{\sqrt{2}du}{\sqrt{1-u^2}}$

$=\sqrt{2}{\sin }^{-1}u+c$

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