Explain solution for  RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 21 maths textbook solution.

Answer:-  $\frac{\pi}{3}$

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

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

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

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

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

Let $\cos x = t$

\begin{aligned} &x=0, t=1 \\ &x=\pi, t=-1 \end{aligned}

\begin{aligned} &2 I=\int_{1}^{-1}-\pi t^{2} d t \\ &I=\frac{-\pi}{2}\left[\frac{t^{3}}{3}\right]_{1}^{-1} \\ &I=\frac{-\pi}{2}\left[\frac{-1}{3}-\frac{1}{3}\right] \end{aligned}

\begin{aligned} &I=\frac{-\pi}{2}\left[\frac{2}{3}\right] \\ &I=\frac{\pi}{3} \end{aligned}