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

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} &00 \rightarrow x \cos (\pi x)>0 \end{aligned}

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

\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}