Discussion Forum

Value of $\int_{0}^{\pi/2} \frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cos x}}dx$ is

Answer:

Explanation:

Let $I =\int_{0}^{\pi/2} \frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cos x}}dx$ ...(1)

Then $I =\int_{0}^{\pi/2} \frac{\sqrt{sin(\pi/2-x)}}{\sqrt{sin(\pi/2-x)}+\sqrt{cos (\pi/2-x)}}dx$

$I =\int_{0}^{\pi/2} \frac{\sqrt{cos x}}{\sqrt{cos x}+\sqrt{sin x}}dx$...(2)

Adding (1) and (2) we get

2I = $\int_{0}^{\pi/2} \frac{\sqrt{sinx}}{\sqrt{cos x}+\sqrt{sinx}}dx$ + $\int_{0}^{\pi/2} \frac{\sqrt{cos x}}{\sqrt{sin x}+\sqrt{cosx}}dx$

= $\int_{0}^{\pi/2} \frac{\sqrt{cos x}+\sqrt{sin x}}{\sqrt{cos x}+\sqrt{sin x}}dx$

= $\int_{0}^{\pi/2}1.dx$=$[x]_0^\left(\pi/2\right)$ = $\pi/2-0$

I = $\pi/4$ = $\int_{0}^{\pi/2} \frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cos x}}dx$ = $\pi/4$