**Toolbox:**

- $\; \int \limits_a^b f(x)dx=F(b)-F(a)$

Step 1:

Given $I= \int \limits_{-1}^{3/2} |x \sin \pi x|dx$

Let $\pi x=t$ on differentiating w.r.t x

$\pi.dx=dt$ Also $x=\pi/t,$ and $dx=dt/\pi$

The limit changes when we substitute t

When $x=-1,t=-\pi$ and

When $4x=3/2,\; t=3 \pi/2$

Therefore $I= \large\frac{1}{\pi^2} $$\int \limits_{-\pi}^{3\pi/2} |\large\frac{t}{\pi}$$ \sin t|\large\frac{dt}{\pi}$

$= \large\frac{1}{\pi^2} $$\int \limits_{-\pi}^{3\pi/2} |t \sin t|dt$

$= \large\frac{1}{\pi^2} $$\int \limits_{-\pi}^{\pi} | t\sin t|dt+ \int \limits_{\pi}^{3 \pi/2} |t\sin t|dt$$

$\sin t$ changes in the 3 rd and 4 th quadrant.

Now let us consider $f(t)=|t \sin t|$ on $\pi,3\pi/2$ (ie) in the 3 rd quadrant

Hence $ t \sin t=-ve$ therefore $|t \sin t|=-t \sin t$ on $[\pi,3\pi/2]$

Also Let $f(t) =|t \sin t|$

Therefore $ f(-t)=|-t. \sin (-t)|= t.\sin t$

Hence it is an even function

Therefore $ \int \limits_{-\pi}^{\pi} |t \sin t|dt= 2 \int \limits_0^{\pi} | t \sin t|dt =2 \int \limits_0^{\pi} t \sin t .dt$

Therefore $ I=\bigg\{ \frac{1}{\pi^2} \bigg( 2\int \limits_0^{\pi} t \sin t dt \bigg)- \int \limits_\pi^{3 \pi/2} t \sin t dt\bigg\}$

Step 2:

Consider $\int t \sin t\;dt$

Clearly this is of the form $\int udv,$ which can be solved by the method of integration

$\int udv=uv-\int vdu$

Let $u=t$ on differentiating w.r.t t,$du=dt$

Let $dv=\sin t dt $ on integrating we get, $v=-\cos t dt$

Therefore $ I= (-t \cos t)-\int -\cos t dt$

$=-(t \cos t)+\int \cos t dt$

On integrating we get

$-( t\cos t)+(\sin t)$

Step 3:

Therefore $ I=\large \frac{1}{\pi^2}$$ \bigg[2 \int \limits_0^{\pi} t \sin t dt -\int \limits_{\pi}^{3\pi/2} t \sin tdt\bigg]$

$= \large\frac{1}{\pi^2} $$\bigg[2( \sin t-t \cos t)\bigg]_0^{\pi}- \bigg[ \sin t- t cos t\bigg]_\pi^{3\pi/2}$

On applying limits

$= \large\frac{1}{\pi^2} $$\bigg[2( \sin \pi-\pi \cos \pi)-0\bigg]-\bigg[\bigg( \sin \large\frac{3\pi}{2}-\frac{3\pi}{2} $$\cos \large\frac{3\pi}{2}\bigg)$$-\bigg(\sin \pi-\pi \cos \pi\bigg)\bigg]$

$\sin \pi=0, \cos \pi=-1, \sin 3\pi/2=-1, \cos 3 \pi/2 =0$

$=\large\frac{1}{\pi^2} $$[2 \pi -(1-\pi)]$

$=\large\frac{1}{\pi^2}$$(3\pi+1)$

$=\large\frac{3}{\pi}+\frac{1}{\pi^2}$