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

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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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