Question
If $$\int\limits_0^\pi {x\,f\left( {\sin \,x} \right)} dx = A\int\limits_0^{\frac{\pi }{2}} {f\left( {\sin \,x} \right)} dx,$$ then $$A$$ is-
A.
$$2\pi $$
B.
$$\pi $$
C.
$$\frac{\pi }{4}$$
D.
$$0$$
Answer :
$$\pi $$
Solution :
$$\eqalign{
& {\text{Let }}I = \int\limits_0^\pi {x\,f\left( {\sin \,x} \right)} dx = \int\limits_0^\pi {\left( {\pi - x} \right)\,f\left( {\sin \,x} \right)} dx \cr
& \therefore 2I = \pi \int\limits_0^\pi {f\left( {\sin \,x} \right)} dx = \pi .2\int\limits_0^{\frac{\pi }{2}} {f\left( {\sin \,x} \right)} dx \cr
& \therefore I = \pi \int\limits_0^{\frac{\pi }{2}} {f\left( {\sin \,x} \right)} dx\,\,\,\,\,\, \Rightarrow A = \pi \cr} $$