The value of the integral \(\rm \displaystyle\int_0^{\pi/2} \dfrac{\sqrt{\sin x}}{\sqrt{\sin x}+ \sqrt{\cos x}}dx\) is

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NIMCET 2013 Official Paper

Answer 1

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 = \(\dfrac{\pi}{4}\)