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  • Explain solution for RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 42 maths textbook solution

Explain solution for  RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 42 maths textbook solution.

Answers (1)

Answer:-  \frac{2}{\pi}+\frac{1}{\pi^{2}}

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

Given:-  \int_{0}^{3 / 2}|x \sin \pi x| \cdot d x

Solution : \int_{0}^{3 / 2}|x \sin \pi x| \cdot d x

               \begin{aligned} &=\int_{0}^{1} x \sin \pi x \cdot d x-\int_{1}^{3 / 2} x \sin \pi x \cdot d x \\ &=\left[\frac{-x \cos x \pi}{\pi}+\frac{\sin \pi x}{\pi^{2}}\right]_{0}^{1}-\left[\frac{-x \cos x \pi}{\pi}+\frac{\sin \pi x}{\pi}\right]_{1}^{3 / 2} \end{aligned}

                \begin{aligned} &=\left[\frac{-1 \cos \pi}{\pi}+\frac{\sin \pi}{\pi^{2}}-0+0\right]-\left[\frac{-3}{2} \frac{\cos ^{3 \pi / 2}}{\pi}+\frac{\cos \pi}{\pi}+\frac{\sin ^{3 \pi / 2}}{\pi^{2}}-\frac{\sin \pi}{\pi}\right]_{1}^{3 / 2} \\ &=\frac{1}{\pi}+\frac{3}{2} \cdot 0+\frac{1}{\pi}+\frac{1}{\pi^{2}}+0 \\ &=\frac{2}{\pi}+\frac{1}{\pi^{2}} \end{aligned}

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