#### Explain solution for  RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 5 maths textbook solution.

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

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

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

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

$\therefore I=\int_{0}^{\pi / 2} \frac{\sin ^{n}\left(\frac{\pi}{2}-x\right)}{\sin ^{n}\left(\frac{\pi}{2}-x\right)+\cos ^{n}\left(\frac{\pi}{2}-x\right)} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right]$

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

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