∫ _ 0 ^Π #160; xsin 2x .dx Apply bypart rule ∫ _ 0 ^Π #160; xsin 2x .dx = 1 / 2 xcos 2x ∫ 1 / 2 cos 2x dx #160; = 1 / 2 ...

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Faraday's Law

∫ 0 π xsin 2x...

Question

$\int_{0}^{π} x sin 2 x . d x$

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Solution

$\int_{0}^{Π} x sin 2 x . d x$
Apply bypart rule
$\int_{0}^{Π} x sin 2 x . d x = - \frac{1}{2} x cos 2 x - \int - \frac{1}{2} cos 2 x d x = - \frac{1}{2} x cos 2 x + \frac{1}{4} sin 2 x$
Now compute the boundary conditions
$= {[- \frac{1}{2} x cos 2 x + \frac{1}{4} sin 2 x]}_{0}^{Π} = - \frac{Π}{2} - 0 = - \frac{Π}{2}$

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