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

Answer:-  $\frac{\pi}{4}$

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

Given:-  $\int_{0}^{\pi / 2} \frac{\tan ^{7} x}{\tan ^{7} x+\cot ^{7} x} d x$

Solution : $I=\int_{0}^{\pi / 2} \frac{\tan ^{7} x}{\tan ^{7} x+\cot ^{7} x} d x$                                                   ....(1)

$I=\int_{0}^{\pi / 2} \frac{\tan ^{7}\left(\frac{\pi}{2}-x\right)}{\tan ^{7}\left(\frac{\pi}{2}-x\right)+\cot ^{7}\left(\frac{\pi}{2}-x\right)} d x$

\begin{aligned} &I=\int_{0}^{\pi / 2} \frac{\cot ^{7} x}{\tan ^{7} x+\cot ^{7} x} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; &...\text { (2) }\left[\because \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right] \end{aligned}

\begin{aligned} &2 I=\int_{0}^{\pi / 2} \frac{\tan ^{7} x+\cot ^{7} x}{\tan ^{7} x+\cot ^{7} x} d x \\ &2 I=\int_{0}^{\pi / 2} 1 \cdot d x \\ &2 I=[x]_{0}^{\pi / 2} \end{aligned}

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