#### Provide solution for  RD Sharma maths Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 32 Subquestion (i) textbook solution.

Answer:-  $\frac{\pi}{2}$

Hints:-  You must know the integral rules of trignometric functions.

Given:-  $\int_{-3 \pi / 2}^{-\pi / 2}\left\{\sin ^{2}(3 \pi+x)+(\pi+x)^{3}\right\} d x$

Solution : $\int_{-3 \pi / 2}^{-\pi / 2}\left\{\sin ^{2}(3 \pi+x)+(\pi+x)^{3}\right\} d x$

$I=\int_{-3 \pi / 2}^{-\pi / 2}\left\{(x+\pi)^{3}+\sin ^{2} x\right\} d x$                                        ....(1)

\begin{aligned} &=\int_{-3 \pi / 2}^{-\pi / 2}\left(\left(\frac{-\pi}{2}-\frac{-3 \pi}{2}-x-\pi\right)^{3}+\sin ^{2}\left(\frac{-\pi}{2}-\frac{-3 \pi}{2}-x\right)\right) d x\\ &=\int_{-3 \pi / 2}^{-\pi / 2}\left(-(x+\pi)^{3}+\sin ^{2} x\right) d x \end{aligned}......(2)

\begin{aligned} &2 I=\int_{-3 \pi / 2}^{-\pi / 2} 2 \sin ^{2} x \cdot d x \\ &2 I=2 \int_{-3 \pi / 2}^{-\pi / 2}\left(\frac{1-\cos 2 x}{2}\right) \cdot d x \\ &2 I=2 \int_{-3 \pi / 2}^{-\pi / 2} \frac{1}{2} \cdot d x-2 \int_{-3 \pi / 2}^{-\pi / 2}\left(\frac{\cos 2 x}{2}\right) \cdot d x \end{aligned}

\begin{aligned} &\left.2 I=x]_{-3 \pi / 2}^{-\pi / 2}-\sin 2 x\right]_{-3 \pi / 2}^{-\pi / 2} \\ &2 I=\frac{-\pi}{2}-\left(\frac{-3 \pi}{2}\right)-\sin 2\left(\frac{-\pi}{2}\right)+\sin 2 x \cdot\left(\frac{-3 \pi}{2}\right) \\ &2 I=\frac{-\pi}{2}+\frac{3 \pi}{2}-\sin (-\pi)+\sin (-3 \pi) \end{aligned}

\begin{aligned} &2 I=\frac{2 \pi}{2}-0 \\ &I=\frac{\pi}{2} \end{aligned}