#### Please solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 (a) Question 4 maths textbook solution.

Answer : $\frac{\pi}{12}$

Given : $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$

Hint : Use the formula of $\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x$

Solution : $I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x \quad-----(1)$

\begin{aligned} &I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin \left(\frac{\pi}{2}-x\right)}}{\sin \left(\frac{\pi}{2}-x\right)+\sqrt{\cos \left(\frac{\pi}{2}-x\right)}} d x \\ &I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x}}{\sqrt{\cos x+\sqrt{\sin x}}} d x------(2) \end{aligned}

\begin{aligned} &2 I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}}{\sqrt{\sin x+\sqrt{\cos x}}} d x+\frac{\sqrt{\cos x}}{\sqrt{\cos x+\sqrt{\sin x}}} d x \\ &2 I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x \\ &2 I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 d x \end{aligned}

\begin{aligned} &2 I=(x)_{\frac{\pi}{6}}^{\frac{\pi}{3}} \\ &2 I=\frac{\pi}{6} \\ &I=\frac{\pi}{12} \end{aligned}