#### Need solution for  RD Sharma maths Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 35 textbook solution.

Answer:-  $\frac{2}{\pi}$

Hints:-  You must know the rules of integration.

Given:-  $\int_{-1}^{1}|x \cdot \cos \pi x| \cdot d x$

Solution : $f(x)=\int_{-1}^{1}|x \cdot \cos \pi x| \cdot d x$

$|f(x)|=f(x) \text { when } x \geq 0 \text { and }-f(x) \text { when } x<0$

$\text { When } f(x)=x \cos (\pi x) \text {, we have }$

\begin{aligned} &|x \cos (\pi x)|=\left\{\begin{array}{l} x \cos (\pi x) ;-1 \leq x \leq 1 / 2 \\ -x \cos (\pi x) ;-1 / 2 \leq x<0 \\ x \cos \pi x ; 0 \leq x<1 / 2 \\ -x \cos (\pi x) ; 1 / 2 \leq x<1 \end{array}\right. \\ &I=\int_{-1}^{1}|x \cdot \cos \pi x| \cdot d x \end{aligned}

$\\I=\int_{-1}^{-1 / 2} x \cdot \cos (\pi x)-\int_{-1 / 2}^{0} x \cdot \cos (\pi x) \cdot d x+\int_{0}^{1 / 2} x \cdot \cos (\pi x) \cdot d x-\int_{-1 / 2}^{1} x \cdot \cos (\pi x) \cdot d x$

Now we can easily compile indefinite integral

\begin{aligned} I &=\int x \cdot \cos (\pi x) \cdot d x \\ &=\frac{x \cdot \cos (\pi x)}{\pi}+\frac{\cos (\pi x)}{\pi} \\ \end{aligned}

$\therefore I^{\prime}=\left(I_{1 / 2}-I_{-1}\right)-\left(I_{0}-I_{-1 / 2}\right)+\left(I_{1 / 2}-I_{0}\right)-\left(I_{1}-I_{1 / 2}\right)$

Where

$\begin{gathered} I_{-1}=\frac{-1}{\pi^{2}} \\ I_{-1 / 2}=\frac{1}{2 \pi} \\ I_{0}=\frac{1}{\pi^{2}} \end{gathered}$

\begin{aligned} &I_{1 / 2}=\frac{1}{2 \pi} \\ &I_{1}=\frac{-1}{\pi^{2}} \end{aligned}

Putting values

We get $I=\frac{2}{\pi}$