Question

Evaluate the integral
${\int }_0^{\pi /2}\frac{\sqrt{cotx}}{\sqrt{tanx}+\sqrt{cotx}}$ $dx$

• A. $\pi$
• B. $\pi /2$
• C. $\pi /3$
• D. $\pi /4$

Answer 1

Answer: (D)

$\pi /4$

$I={\int }_0^{\frac{\pi }{2}}\frac{\sqrt{\cot x}}{\sqrt{\tan x}+\sqrt{\cot x}}dx$

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

$I={\int }_0^{\frac{\pi }{2}}\frac{\sqrt{\cot x}}{\sqrt{\tan x}+\sqrt{\cot x}}dx={\int }_0^{\frac{\pi }{2}}\frac{\sqrt{\tan x}}{\sqrt{\cot x}+\sqrt{\tan x}}dx+I$

$2I={\int }_0^{\frac{\pi }{2}}1\cdot dx$

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