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  • Explain solution for RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 40 maths textbook solution.

Explain solution for  RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 40 maths textbook solution.

Answers (1)

Answer:-  \frac{1}{\pi}\left(\frac{5}{2}-\frac{1}{\pi}\right)

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

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

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

               \begin{aligned} &0<x<\frac{1}{2} \\ &0<\pi x<\frac{\pi}{2} \\ &\cos \pi x>0 \rightarrow x \cos (\pi x)>0 \end{aligned}

               \begin{aligned} &|x \cos \pi x|=x \cos \pi x \\ &\frac{1}{2}<x<\frac{3}{2} \end{aligned}

                 \begin{aligned} &\frac{\pi}{2}<\pi x<\frac{3 \pi}{2} \\ &\cos \pi x<0 \rightarrow x \cos \pi x<0 \\ &|x \cos \pi x|=-x \cos \pi x \end{aligned}

                \begin{aligned} &I=\int_{0}^{3 / 2}|x \cos \pi x| \cdot d x \\ &I=\int_{0}^{3 / 2} x \cos \pi x+\int_{1 / 2}^{3 / 2}-(x \cos \pi x) \\ &I=\int_{0}^{1 / 2} x \cos \pi x-\int_{1 / 2}^{3 / 2} x \cos \pi x \cdot d x \end{aligned}

               \begin{aligned} &\int x \cos \pi x=x \frac{\sin \pi x}{\pi}-\int \frac{\sin \pi x}{\pi} \\ &\int x \cos \pi x=\frac{x}{\pi} \sin \pi x+\frac{\cos \pi x}{\pi^{2}} \end{aligned}

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

              \begin{aligned} &=\left[\frac{1}{2 \pi}-\frac{1}{\pi^{2}}\right]-\left[\frac{-2}{\pi}\right] \\ &=\left[\frac{5}{2 \pi}-\frac{1}{\pi^{2}}\right] \\ &=\frac{1}{\pi}\left[\frac{5}{2}-\frac{1}{\pi}\right] \end{aligned}

           

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