I = \(\int\limits^{\frac{\pi}{2}}_0\)\(\frac{\sqrt{tanx}}{\sqrt{tanx + cotx}}\)dx .......... (1)
I = \(\int\limits^{\frac{\pi}{2}}_0\)\(\frac{\sqrt{tan(\frac{\pi}{2}-x)}}{\sqrt{tan(\frac{\pi}{2}-x)} + \sqrt{cotx(\frac{\pi}{2}-x)}} dx\) (\(\because \)\(\int\limits^{b}_a\)f(x)dx = \(\int\limits^{b}_a\)f(a+b - x)dx)
= \(\int\limits^{\frac{\pi}{2}}_0\)\(\frac{\sqrt{cot x}} {\sqrt{cotx} +\sqrt{tan x}} dx\) ......... (2)
By adding equations (1) and (2), we get
I = \(\int\limits^{\frac{\pi}{2}}_0\) \(\frac{\sqrt{tan x}+ \sqrt{cot x}} {\sqrt{tan x} +\sqrt{cot x}} dx\)
= \(\int\limits^{\frac{\pi}{2}}_0\) dx = \([x]^\frac{\pi}{2}_{0}\)
= \(\frac{\pi}{2} - 0\) = \(\frac{\pi}{2}\)
\(\Rightarrow\) I = \(\frac{\pi}{4}\)
Hence, \(\int\limits^{\frac{\pi}{2}}_0\)\(\frac{\sqrt{tanx}}{\sqrt{tanx}+{\sqrt{cot x}}}\) = \(\frac{\pi}{4}\)