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Question

$_{0}^{π / 2} \int x sin x d x$ .

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Solution

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$\int_{0}^{\frac{π}{2}} x sin x d x$
$= {[- x cos x]}_{0}^{\frac{π}{2}} + \int_{0}^{\frac{π}{2}} cos x d x$
$= {[- x cos x + sin x]}_{0}^{\frac{π}{2}}$
$= [1 - 0]$
$= 1$

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