Question

If ${\int }_0^{\lambda }xf(\sin x)dx=A{\int }_0^{\pi /2}f(\sin x)dx,$ then A is equal to

• A. $0$
• B. $\pi$
• C. $\frac{\pi }{4}$
• D. $2\pi$
• E. $3\pi$

Answer 1

Answer: (B)

Let $I={\int }_0^{\pi }xf(\sin x)dx$ ...(i)
$={\int }_0^{\pi }(\pi -x)f[\sin (\pi -x)]dx$
$\Rightarrow$ $={\int }_0^{\pi }(\pi -x)f(\sin x)dx$ ...(ii)
On adding Eqs. (i) and (ii), we get
$2I={\int }_0^{\pi }\pi f(\sin x)dx$
$\Rightarrow$ $I=\frac{\pi }{2}{\int }_0^{\pi }f(\sin x)dx=\pi {\int }_0^{\pi /2}f(\sin x)dx$
$\Rightarrow$ $A{\int }_0^{\pi /2}f(\sin x)dx=\pi {\int }_0^{\pi /2}f(\sin x)dx$
$\Rightarrow$ $A=\pi$