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Q.
If $I_1$ = $\int\limits_0^ {\pi/2}\, x Sin \: x \: dx$ and $I_2=\int\limits_0^ {\pi/2} \: x \: Cos \: x \: dx$ then which one of the following is true ?
Since,$ I_{1}=\int\limits_{0}^{\pi / 2} x \sin x \,d x $
and $ I_{2}=\int\limits_{0}^{\pi / 2} x \cos x \,d x $
$\therefore I_{1}=\int\limits_{0}^{\pi / 2} x \sin x \,d x$
$\Rightarrow I_{1} =-[x \cos x]_{0}^{\pi / 2}-\int_{0}^{\pi / 2}-\cos x d x $
$=[0-0]+[\sin x]_{0}^{\pi / 2}$
$=1 \,\,\,\,\,\dots(i)$
and $ I_{2} =\int\limits_{0}^{\pi / 2} x \cos x \,d x $
$=[x \sin x]_{0}^{\pi / 2}-\int_{0}^{\pi / 2} \cos \,x d x$
$=\left(\frac{\pi}{2}-0\right)+[\sin x]_{0}^{\pi / 2} $
$=\frac{\pi}{2}+1$
From (i) and (ii)
$I_{2}-I_{1}=\frac{\pi}{2}$