Correct Answer - Option 4 :
\(\dfrac{\pi}{4}\)
Concept:
\(\rm \displaystyle\int_{a}^{b} f(x) dx = \displaystyle\int_{a}^{b} f(a+b-x) dx\)
Calculations:
Consider, I = \(\rm \displaystyle\int_0^{\pi/2} \dfrac{\sqrt{\sin x}}{\sqrt{\sin x}+ \sqrt{\cos x}}dx\) ....(1)
I = \(\rm \displaystyle\int_{0}^{\pi/2} \dfrac{\sqrt{\sin (\dfrac{\pi}{2}-x)}}{\sqrt{\sin (\dfrac{\pi}{2}-x)}+ \sqrt{\cos (\dfrac{\pi}{2}-x)}}dx\)
I = \(\rm \displaystyle\int_0^{\pi/2} \dfrac{\sqrt{\cos x}}{\sqrt{\cos x}+ \sqrt{\sin x}}dx\) ....(2)
Adding (1) and (2), we have
2I = \(\rm \displaystyle\int_0^{\pi/2} \dfrac{\sqrt{\cos x}+ \sqrt{sinx}}{\sqrt{\cos x}+ \sqrt{\sin x}}dx\)
2I = \(\rm \displaystyle\int_0^{\pi/2}dx\)
2I = \(\rm[x]^\frac{\pi}{2}_0\)
I = \(\dfrac{\pi}{4}\)
