Question

If $I_1$ = $\underset{0}{\overset{\pi /2}{\int }}xSinxdx$ and $I_2=\underset{0}{\overset{\pi /2}{\int }}xCosxdx$ then which one of the following is true ?

• A. $I_1=I_2$
• B. $I_1+I_2$
• C. $I_1=\frac{\pi }{2}{I}_2$
• D. $I_1-I_2=\frac{\pi }{2}$

Answer 1

Answer: (D)

Since,$I_1=\underset{0}{\overset{\pi /2}{\int }}x\sin xdx$
and $I_2=\underset{0}{\overset{\pi /2}{\int }}x\cos xdx$
$\therefore I_1=\underset{0}{\overset{\pi /2}{\int }}x\sin xdx$
$\Rightarrow I_1=-[x\cos x{]}_0^{\pi /2}-{\int }_0^{\pi /2}-\cos xdx$
$=[0-0]+[\sin x{]}_0^{\pi /2}$
$=1\dots (i)$
and $I_2=\underset{0}{\overset{\pi /2}{\int }}x\cos xdx$
$=[x\sin x{]}_0^{\pi /2}-{\int }_0^{\pi /2}\cos xdx$
$=\left(\frac{\pi }{2}-0\right)+[\sin x{]}_0^{\pi /2}$
$=\frac{\pi }{2}+1$
From (i) and (ii)
$I_2-I_1=\frac{\pi }{2}$