Prove that: ^ _0xf( )dx= 2 ^ _0f( )dx - Vidyadip

Question

Prove that:
$\int\limits^\pi_0\text{xf}(\sin\text{x})\text{dx}=\frac{\pi}{2}\int\limits^\pi_0\text{f}(\sin\text{x})\text{dx}$

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Answer

$\int\limits^\pi_0\text{xf}(\sin\text{x})\text{dx}=\int\limits^\pi_0(\pi-\text{x})\text{f}\big[\sin(\pi-\text{x})\big]\text{dx}$ $\Bigg[\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\text{f}(\text{a}-\text{x})\text{dx}\Bigg]$
$\Rightarrow\int\limits^\pi_0\text{xf}(\sin\text{x})\text{dx}=\int\limits^\pi_0(\pi-\text{x})\text{f}(\sin\text{x})\text{dx}$
$\Rightarrow\int\limits^\pi_0\text{xf}(\sin\text{x})\text{dx}=\pi\int\limits^\pi_0\text{f}(\sin\text{x})\text{dx}-\int\limits^\pi_0\text{xf}(\sin\text{x})\text{dx}$
$\Rightarrow2\int\limits^\pi_0\text{xf}(\sin\text{x})\text{dx}=\pi\int\limits^\pi_0\text{f}(\sin\text{x})\text{dx}$
$\Rightarrow\int\limits^\pi_0\text{xf}(\sin\text{x})\text{dx}=\frac{\pi}{2}\int\limits^\pi_0\text{f}(\sin\text{x})\text{dx}$

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