Question

${\int }_0^{\frac{\pi }{4}}\left(\sqrt{\tan x}+\sqrt{\cot x}\right)dx$ is equal to

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

Answer 1

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

${\int }_0^{\frac{\pi }{4}}(\sqrt{\tan x}+\sqrt{\cot x})dx$

$={\int }_0^{\frac{\pi }{4}}(\sqrt{\frac{\sin x}{\cos x}}+\sqrt{\frac{\cos x}{\sin x}})dx$

$={\int }_0^{\frac{\pi }{4}}\left(\frac{\sin x+\cos x}{\sqrt{\sin x\cos x}}\right)dx$

$=\sqrt{2}{\int }_0^{\frac{\pi }{4}}\left(\frac{\sin x+\cos x}{\sqrt{2\sin x\cos x}}\right)dx$

$=\sqrt{2}{\int }_0^{\frac{\pi }{4}}\left(\frac{\sin x+\cos x}{\sqrt{1-{(\sin x-\cos x)}^2}}\right)dx$

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

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

When $x=0\Rightarrow t=-1$

When $x=\frac{\pi }{4}\Rightarrow t=0$

$\therefore I=\sqrt{2}{\int }_{-1}^0\frac{dt}{\sqrt{1-t^2}}$

$=\sqrt{2}{\left[{\sin }^{-1}t\right]}_{-1}^0$

$=\sqrt{2}[{\sin }^{-1}0-{\sin }^{-1}(-1)]$

$=\sqrt{2}[0+\frac{\pi }{2}]=\sqrt{2}\times \frac{\pi }{2}\times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\pi }{\sqrt{2}}$

$\therefore I=\frac{\pi }{\sqrt{2}}$