#### Please solve RD Sharma class 12 chapter 19 Definite Integrals exercise Multiple choice question 5 maths textbook solution

$\frac{\pi }{4}$
Given:

$\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x$

Hint:

using $\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x$

Explanation:

Let

$I=\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x \ldots(i)$

$I=\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\cos \left(\frac{\pi}{2}-x\right)}}{\sqrt{\cos \left(\frac{\pi}{2}-x\right)}+\sqrt{\sin \left(\frac{\pi}{2}-x\right)}} d x, \quad\left[\therefore \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right]$$=\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x \ldots(i i)$

\begin{aligned} &2 I=\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\cos x}+\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x \\\\ &=\int_{0}^{\frac{\pi}{2}} 1 d x \\\\ &2 I=[x]_{0}^{\frac{\pi}{2}} \\\\ &2 I=\left[\frac{\pi}{2}-0\right] \end{aligned}

\begin{aligned} &2 I=\frac{\pi}{2} \\\\ &I=\frac{\pi}{4} \end{aligned}