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Explain solution for  RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 41 maths textbook solution.

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Answer:-  \frac{1}{\pi}

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

Given:-  \int_{0}^{1}|x \sin \pi x| d x

Solution : \int_{0}^{1}|x \sin \pi x| d x

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

               \begin{aligned} &=\left[(-1) \frac{-1}{\pi}+0\right] \\ &=\frac{1}{\pi} \end{aligned}

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