_0^ xf ( x )dx = - Vidyadip

MCQ

$\mathop \smallint \limits_0^\pi xf\left( {\sin x} \right)dx = $

A
$\;\pi \mathop \smallint \limits_0^\pi xf\left( {\cos x} \right)dx$
B
$\;\pi \mathop \smallint \limits_0^\pi f\left( {\sin x} \right)dx$
C
$\frac{\pi }{2}\mathop \smallint \limits_0^{\frac{\pi }{2}} f\left( {\sin x} \right)dx$
✓
$\pi \mathop \smallint \limits_0^{\frac{\pi }{2}} f\left( {\cos x} \right)dx$

✓

Answer

Correct option: D.

$\pi \mathop \smallint \limits_0^{\frac{\pi }{2}} f\left( {\cos x} \right)dx$

d
$I=\int_{0}^{\pi} x f(\sin x) d x$

$=\int_{0}^{\pi}(\pi-x) f(\sin x) d x$

$=\pi \int_{0}^{\pi} f(\sin x) d x-1$

$\Rightarrow 2 I=\pi \frac{\pi}{0} f(\sin x) d x$

$I=\frac{\pi}{2} \int_{0}^{\pi} f(\sin x) d x$

$=\pi \int_{0}^{\pi / 2} f(\sin x) d x$

$=\pi \int_{0}^{\pi / 2} f(\cos x) d x$

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