Question

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

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

Answer 1

Answer: (D)

Let $I =\int_0^\pi xf (\sin x ) dx \ldots(1)$
$I=\int_0^\pi(\pi-x) f(\sin (\pi-x)) d x=\int_0^\pi(\pi-x) f(\sin x) d x$...(2)
Adding (1) and (2)
$2 I=\int_0^\pi(x+\pi-x) f(\sin x) d x=\pi \int_0^\pi f(\sin x) d x $
$\Rightarrow 2 I=2 \pi \int_0^{\pi / 2} f(\sin x) d x \Rightarrow I=\pi \int_0^{\pi / 2} f(\sin x) d x $
$\Rightarrow A=\pi$