Question

If $I_1={\int }_0^{\pi /2}x\sin xdx$ and $I_2={\int }_0^{\pi /2}x\cos xdx$, then which one of the following is true?

• A. $I_1+I_2=\frac{\pi }{2}$
• B. $I_2-I_1=\frac{\pi }{2}$
• C. $I_1+I_2=0$
• D. $I_1=I_2$

Answer 1

The correct option is B $I_2-I_1=\frac{\pi }{2}$

Since, $I_1={\int }_0^{\pi /2}x\sin xdx$
and $I_2={\int }_0^{\pi /2}x\cos xdx$
Therefore, $I_1={\int }_0^{\pi /2}x\sin xdx$
$\Rightarrow I_1=-{\left[x\cos x\right]}_0^{\pi /2}-{\int }_0^{\pi /2}-\cos xdx$
$=[0-0]+{\left[\sin x\right]}_0^{\pi /2}$
$=1$ ....(i)
and $I_2={\int }_0^{\pi /2}x\cos xdx$
$={\left[x\sin x\right]}_0^{\pi /2}-{\int }_0^{\pi /2}\sin xdx$
$=(\frac{\pi }{2}-0)+{\left[\sin x\right]}_0^{\pi /2}$
$=\frac{\pi }{2}+1$ ...(ii)
From equations (i) and (ii), we get
$I_2-I_1=\frac{\pi }{2}$