Question

${\int }_0^{\pi /2}\frac{dx}{1+\sqrt{\tan x}}$ equals

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{8}$
• D. 0

Answer 1

The correct option is A $\frac{\pi }{4}$

$I={\int }_0^{\pi /2}\frac{dx}{1+\sqrt{\tan x}}={\int }_0^{\pi /2}\frac{\sqrt{\cos x}dx}{\sqrt{\sin x}+\sqrt{\cos x}}$ ......... (i)

$I={\int }_0^{\pi /2}\frac{\sqrt{\cos (\frac{\pi }{2}-x)}}{\sqrt{\sin (\frac{\pi }{2}-x)}+\sqrt{\cos (\frac{\pi }{2}-x)}}dx$

$\because {\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx$

$\therefore I={\int }_0^{\pi /2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx$ .......... (ii)

Adding (i) and (ii) we get,

$2I={\int }_0^{\pi /2}\frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx={\int }_0^{\pi /2}dx=\frac{\pi }{2}$

or $I=\frac{\pi }{4}$.