Mathematics

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

##### SOLUTION
Let  $I=\displaystyle\int_{0}^{\pi}{xf\left(\sin{x}\right)dx}$       ........$(1)$
Replace $x\rightarrow 0+\pi-x=\pi-x$
$\Rightarrow I=\displaystyle\int_{0}^{\pi}{\left(\pi-x\right)f\left(\sin{\left(\pi-x\right)}\right)dx}$
$\Rightarrow I=\displaystyle\int_{0}^{\pi}{\left(\pi-x\right)f\left(\sin{x}\right)dx}$.....$(2)$
Adding $(1)$ and $(2)$ we get
$2I=\displaystyle\int_{0}^{\pi}{xf\left(\sin{x}\right)+\left(\pi-x\right)f\left(\sin{x}\right)dx}$
$2I=\displaystyle\int_{0}^{\pi}{\left[xf\left(\sin{x}\right)+\pi f\left(\sin{x}\right)-xf\left(\sin{x}\right)\right]dx}$
$2I=\displaystyle\int_{0}^{\pi}{\left[\pi f\left(\sin{x}\right)\right]dx}$
$2I=\pi\displaystyle\int_{0}^{\pi}{\left[f\left(\sin{x}\right)\right]dx}$
$2I=2\pi\displaystyle\int_{0}^{\frac{\pi}{2}}{\left[f\left(\sin{x}\right)\right]dx}$ since $\displaystyle\int_{0}^{2a}{f\left(x\right)dx}=2\displaystyle\int_{0}^{a}{f\left(x\right)}$ if $f\left(2a-x\right)=f\left(x\right)$
$\therefore I=\pi\displaystyle\int_{0}^{\frac{\pi}{2}}{f\left(\sin{x}\right)dx}$
$=\pi\displaystyle\int_{0}^{\frac{\pi}{2}}{f\left(\sin{\left(\dfrac{\pi}{2}-x\right)}\right)dx}$
$\therefore I=\pi\displaystyle\int_{0}^{\frac{\pi}{2}}{f\left(\cos{x}\right)dx}$

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

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