Mathematics

# $\underset{0}{\overset{\pi}{\int}} x \, f (\sin \, x) dx =$

$\pi \underset{0}{\overset{\pi}{\int}} f (\sin \, x) dx$

##### SOLUTION
$\displaystyle\int_{0}^{\pi } xf(\sin x)dx$

$=\displaystyle\int_{0}^{\pi } f(\sin x)dx\left [ \int_{0}^{\pi }x \sin x dx \right ]$

$=\displaystyle\int_{0}^{\pi } f(\sin x)dx\left [ -x \cos x -\int_{0}^{\pi}-\cos x dx \right ]$

$=\displaystyle\int_{0}^{\pi } f(\sin x)dx\left [ -x \cos x +sinx \right ]_0^{\pi}$

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

$=\displaystyle\pi \int_{0}^{\pi } f(\sin x)dx$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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