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
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