MCQ
જો ${I_1} = \int_a^{\pi - a} {xf(\sin x)dx,\,{I_2} = \int_a^{\pi - a} {\,\,f(\sin x)dx} } $, તો ${I_2} = . . .$
- A$\frac{\pi }{2}{I_1}$
- B$\pi \,{I_1}$
- ✓$\frac{2}{\pi }{I_1}$
- D$2{I_1}$
$ = \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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