# Evaluate:$\int \limits_{-1}^{3/2} | x \sin \pi x|dx$

Toolbox:
• $(i)\; \int \limits_a^b f(x)dx=F(b)-F(a)$
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)$
$=\large\frac{3}{\pi}+\frac{1}{\pi^2}$