Question

Evaluate: ${\int }_0^{\frac{\pi }{4}}[\sqrt{\tan x}+\sqrt{\cot x}]dx$

• A. $\frac{\pi }{\sqrt{2}}$
• B. $\frac{\pi }{2}$
• C. $\frac{3\pi }{\sqrt{2}}$
• D. $\pi$

Answer 1

The correct option is A $\frac{\pi }{\sqrt{2}}$

${\int }_0^{\frac{\pi }{4}}\frac{\sin x+\cos x}{\sqrt{\sin x\cos x}}dx$
Put $\sin x-\cos x=t\Rightarrow (\sin x+\cos x)dx=dt$
$1-2\sin x\cos x=t^2\Rightarrow \sin x\cos x=\frac{1-t^2}{2}$
${\int }_{-1}^0\frac{1}{\sqrt{\frac{1-t^2}{2}}}dt=\sqrt{2}{\left[{\sin }^{-1}t\right]}_{-1}^0=\frac{\pi }{\sqrt{2}}$