Question

${\int }_0^{\frac{\pi }{2}}\frac{\sqrt{cotx}}{\sqrt{cotx}+\sqrt{tanx}}dx=$ [MP PET 1990, 95; IIT 1983; MNR 1990]

• A. $\pi$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{3}$

Answer 1

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

${\int }_0^{\frac{\pi }{2}}\frac{\sqrt{cotx}}{\sqrt{cotx}+\sqrt{tanx}}dx\cdots \left(i\right)$
$={\int }_0^{\frac{\pi }{2}}\frac{\sqrt{cot(\frac{\pi }{2}-x)}}{\sqrt{cot(\frac{\pi }{2}-x)}+\sqrt{tan(\frac{\pi }{2}-x)}}dx$
$={\int }_0^{\frac{\pi }{2}}\frac{\sqrt{tanx}}{\sqrt{tanx}+\sqrt{cotx}}dx\cdots \left(ii\right)$
Now adding (i) and (ii), we get
$2I={\int }_0^{\frac{\pi }{2}}\frac{\sqrt{\cot x}+\sqrt{tanx}}{\sqrt{tanx}+\sqrt{cotx}}dx=[x{]}_0^{\frac{\pi }{2}}\Rightarrow I=\frac{\pi }{4}$