#### Need solution for RD Sharma maths Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 17 textbook solution.

Answer:- $\frac{\pi^{2}}{4}$

Hints:-  You must know about the integration rules of trigonometry functions.

Given:-  $\int_{0}^{\pi} x \cos ^{2} x d x$

Solution :  $\int_{0}^{\pi} x \cos ^{2} x d x=I$

\begin{aligned} &I=\int_{0}^{\pi}(\pi-x) \cos ^{2}(\pi-x) d x \\ &I=\int_{0}^{\pi}(\pi-x) \cos ^{2} x \cdot d x \\ &I=\int_{0}^{\pi} \pi \cos ^{2} x \cdot d x-\int_{0}^{\pi} x \cos ^{2} x \cdot d x \end{aligned}

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

$\begin{gathered} =\pi\left[\frac{1}{2}\left[\frac{\sin 2 x}{2}\right]_{0}^{\pi}+\frac{1}{2}[x]_{0}^{\pi}\right] \\ 2 I=\frac{\pi}{4}[\sin 2 \pi-\sin 0]+\frac{\pi}{2}[\pi] \end{gathered}$

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