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

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

Answers (1)

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)

Add (1) and (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}

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