Question

Value of $\int\limits_{0}^{\pi /2} \frac{\sqrt{sin x}}{\sqrt{sin x}\sqrt{cos x}} $ dx is

• A. $\frac{\pi}{2}$
• B. $\frac{-\pi}{2}$
• C. $\frac{\pi}{4}$
• D. None of these

Answer 1

The Correct Option is C

Let I= $\int\limits_{0}^{\pi /2} \frac{\sqrt{sin x}}{\sqrt{sin x}+\sqrt{cos x}} dx \quad\ldots\left(i\right)$ $Then, I $= $\int\limits_{0}^{\pi/ 2} \frac{\sqrt{sin \left(\pi /2 -x\right)}}{\sqrt{sin \left(\pi /2-x\right)}+\sqrt{cos \left(\pi /2-x\right)}} dx$ $\Rightarrow\quad I=\int\limits_{0}^{\pi /2} \frac{\sqrt{cos x}}{\sqrt{cos x}+\sqrt{sin x}} dx ...\left(ii\right)$ Adding (i) and (ii), we get 21 $\int\limits_{0}^{\pi /2} \frac{\sqrt{sin \, x}}{\sqrt{cos \,x}+\sqrt{sin\, x}} dx +\int\limits_{0}^{\pi/ 2} \frac{\sqrt{cos \, x}}{\sqrt{sin\,x}+\sqrt{cos\, x}}dx $ $\int\limits_{0}^{\pi /2} \frac{\sqrt{sin\, x}+\sqrt{cos\, x}}{\sqrt{sin \, x}+\sqrt{cos \, x}} dx =\int\limits_{0}^{\pi /2} 1. d x =\left[x\right]_{0}^{\pi /2}=\frac{\pi}{2}-0 $ $\Rightarrow\quad I=\frac{\pi}{4} \Rightarrow \int\limits_{0}^{\pi /2} \frac{\sqrt{sin \,x}}{\sqrt{sin \, x}+\sqrt{cos\, x}} dx=\frac{\pi}{4}$