MCQ
જો $\int_0^\pi {xf(\sin x)dx = A} \int_0^{\pi /2} {f(\sin x)dx} $, તો $A=$
- A$2\pi $
- ✓$\pi $
- C$\frac{\pi }{4}$
- D$0$
Now, $2I = \int_0^\pi {xf(\sin x)dx + \int_0^\pi {(\pi - x)f[\sin (\pi - x)]dx} } $
$ = \int_0^\pi {\pi f(\sin x)dx} = \pi \int_0^\pi {f(\sin x)dx} $
==> $2I = 2\pi \int_0^{\pi /2} {f(\sin x)dx} $
$\therefore$ $I = \pi \int_0^{\pi /2} {f(\sin x)dx} $
$ = A\int_0^\pi {f(\sin x)dx} $.
Hence $A = \pi $.
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