Question

Evaluate: ${\int }_{-1}^{1}t\cos tdt$

Answer 1

intergrating by parts

$I={\int }_{-1}^{1}t\cos tdt$ , where $t=I$ function $\cos t=II$ function

$=t\int \cos tdt-\int \frac{d\left(t\right)}{dt}\int \cos tdt$

$=[t\sin t+\cos t]$

Now putting the values of limit

$I=[t\sin t+\cos t{]}_{-1}^1$

$=(\sin 1+\cos 1)-(-\sin (-1)+\cos (-1\left)\right)$

Now since $\sin (-x)=-\sin x$

and $\cos (-x)=\cos x$

$\therefore I=\sin 1+\cos 1-(\sin 1+\cos 1)$

$=0$