#### Provide solution for  RD Sharma maths Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 13 textbook solution.

Answer:-  $\frac{2\pi}{3}$

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

Given:-  $\int_{0}^{\pi} x \sin ^{3} x \cdot d x$

Solution :  $\int_{0}^{\pi} x \sin ^{3} x \cdot d x=I$

$I=\int_{0}^{\pi} x \sin ^{3} x . d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right]$

\begin{aligned} &I=\int_{0}^{\pi}(\pi-x)[\sin (\pi-x)]^{3} \cdot d x \\ &I=\int_{0}^{\pi}(\pi-x) \sin ^{3} x \cdot d x \\ &I=\int_{0}^{\pi} \pi \sin ^{3} x \cdot d x-\int_{0}^{\pi} x \cdot \sin ^{3} x \cdot d x \end{aligned}

\begin{aligned} &I=\pi \int_{0}^{\pi} \sin ^{3} x \cdot d x-I \\ &2 I=\pi \int_{0}^{\pi} \sin ^{3} x \cdot d x \\ &2 I=\pi \int_{0}^{\pi} \frac{1}{4}(3 \sin x-\sin 3 x) \cdot d x \end{aligned}

\begin{aligned} &I=\frac{\pi}{8} \int_{0}^{\pi} 3 \sin x-\sin 3 x \cdot d x \\ &I=\frac{\pi}{8}\left[3(-\cos x)+\frac{\cos 3 x}{3}\right]_{0}^{\pi} \end{aligned}

Putting limits

\begin{aligned} &I=\frac{\pi}{8}\left[-3 \cos \pi+\frac{\cos 3 x}{3}\right]-\left[-3 \cos 0+\frac{\cos 3(0)}{3}\right] \\ &I=\frac{\pi}{8}\left[(-3)(-1)+\frac{(-1)}{3}\right]-\left[-3+\frac{1}{3}\right] \\ &I=\frac{\pi}{8} \times 2\left[3-\frac{1}{3}\right] \end{aligned}

\begin{aligned} &I=\frac{\pi}{4}\left[\frac{9-1}{3}\right] \\ &I=\frac{\pi}{4} \times \frac{8}{3} \\ &=\frac{2 \pi}{3} \end{aligned}