Answer
$\displaystyle -\frac{π}{4} ≈ -0.785398$
Work Step by Step
Integration by Parts formula: $\displaystyle \int fg' $ $\displaystyle dx = fg-\int f'g$ $dx$
Double Angle Theorem:$\quad \sin(2x) = 2\sin(x)\cos(x)\quad \displaystyle \rightarrow\quad\frac{1}{2}\sin(2x) = \sin(x)\cos(x)$
$\displaystyle \int_0^πx\cos(x)\sin(x)dx = \int_0^πx[\frac{1}{2}\sin(2x)]dx = \frac{1}{2}\int_0^πx[\sin(2x)]dx$
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Let $\displaystyle f = x$ and $\displaystyle g' = \sin(2x)$
thus $\displaystyle f' = 1$ and $\displaystyle g = -\frac{1}{2}\cos(2x)$
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$\displaystyle \frac{1}{2}\int_0^πx[\sin(2x)]dx = \frac{1}{2}[(x[-\frac{1}{2}\cos(2x)])|_0^π - \int_0^π(1)[-\frac{1}{2}\cos(2x)]dx]$
$\displaystyle \frac{1}{2}([-\frac{x}{2}\cos(2x)]_0^π - [-\frac{1}{4}\sin(2x)]_0^π)$
$\displaystyle \frac{1}{2}([-\frac{π}{2}\cos(2π) - (-\frac{0}{2}\cos(0))] - [-\frac{1}{4}\sin(2π) - (-\frac{1}{4}\sin(0))])$
$\displaystyle \frac{1}{2}([-\frac{π}{2}(1) - 0] - [-\frac{1}{4}(0) + \frac{1}{4}(0)])$
$\displaystyle \frac{1}{2}(-\frac{π}{2} + 0)$
$\displaystyle -\frac{π}{4} ≈ -0.785398$
