Let I_1 = _a^ - a xf( x)dx, , I_2 = _a^ - a , ,f( x)dx , then I_2… - Vidyadip

MCQ

Let ${I_1} = \int_a^{\pi - a} {xf(\sin x)dx,\,{I_2} = \int_a^{\pi - a} {\,\,f(\sin x)dx} } $, then ${I_2}$ is equal to

A
$\frac{\pi }{2}{I_1}$
B
$\pi \,{I_1}$
✓
$\frac{2}{\pi }{I_1}$
D
$2{I_1}$

✓

Answer

Correct option: C.

$\frac{2}{\pi }{I_1}$

c
(c) ${I_1} = \int_a^{\pi - a} {xf(\sin x)dx} $

$ = \int_a^{\pi - a} {(\pi - x)\,f\,(\sin (\pi - x))\,dx} $,

$[ \because \int_a^b {f(x)dx = \int_a^b {f(a + b - x)\,dx} } ]$

$ = \int_a^{\pi - a} {(\pi - x)\,f\,(\sin x)\,dx} $

$ = \int_a^{\pi - a} {\pi \,f\,(\sin x)\,dx - {I_1}} $

$ \Rightarrow 2{I_1} = \pi \,{I_2}\, $

$\Rightarrow {I_2} = \frac{2}{\pi }{I_1}$.

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