#### Explain solution for  RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 4 maths textbook solution.

Answer:- $\frac{\pi}{4}$

Hints:-  You must know the integration rules of trigonometric functions and its limits

Given:-  $\int_{0}^{\pi / 2} \frac{\sin ^{3 / 2} x}{\sin ^{3 / 2} x+\cos ^{3 / 2} x} d x$

Solution :   $I=\int_{0}^{\pi / 2} \frac{\sin ^{3 / 2} x}{\cos ^{3 / 2} x+\sin ^{3 / 2} x} d x$                                           ....(1)

\begin{aligned} &I=\int_{0}^{\pi / 2} \frac{\sin ^{3 / 2}\left(\frac{\pi}{2}-x\right)}{\cos ^{3 / 2}\left(\frac{\pi}{2}-x\right)+\sin ^{3 / 2}\left(\frac{\pi}{2}-x\right)} d x\\ &\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \left[\because \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right]\\ &I=\int_{0}^{\pi / 2} \frac{\cos ^{3 / 2} x}{\sin ^{3 / 2} x+\cos ^{3 / 2} x} d x \end{aligned}

\begin{aligned} &I+I=\int_{0}^{\pi / 2} \frac{\sin ^{3 / 2} x}{\sin ^{3 / 2} x+\cos ^{3 / 2} x} d x+\int_{0}^{\pi / 2} \frac{\cos ^{3 / 2} x}{\sin ^{3 / 2} x+\cos ^{3 / 2} x} d x \\ &2 I=\int_{0}^{\pi / 2} \frac{\sin ^{3 / 2} x+\cos ^{3 / 2} x}{\sin ^{3 / 2} x+\cos ^{3 / 2} x} d x \\ &2 I=\int_{0}^{\pi / 2} 1 . d x \end{aligned}

\begin{aligned} &2 I=[x]_{0}^{\pi / 2} \\ &2 I=\left[\frac{\pi}{2}-0\right] \\ &I=\frac{\pi}{4} \end{aligned}