Answer
$-2\pi^2 ≈ -19.739208$
Work Step by Step
Integration by parts formula: $\displaystyle \int fg' $ $\displaystyle dx = fg-\int f'g$ $dx$
$\displaystyle \int_0^{2π}t^2\sin(2t)dt$
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Let $\displaystyle f = t^2$ and $\displaystyle g' = \sin(2t)$
thus $\displaystyle f' = 2t$ and $\displaystyle g = -\frac{1}{2}\cos(2t)$
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$\displaystyle \int_0^{2π}t^2\sin(2t)dt = [t^2(-\frac{1}{2}\cos(2t))]_0^{2\pi} - \int_0^{2\pi}2t(\frac{1}{2}\cos(2t))dt$
$\displaystyle [-\frac{t^2}{2}\cos(2t)]_0^{2\pi} - ([2t(-\frac{1}{4}\sin(2t))]_0^{2π}-\int_0^{2π}2(-\frac{1}{4}\sin(2t))dt)$
$\displaystyle [-\frac{4\pi^2}{2}\cos(4\pi) - (-\frac{0}{2}\cos(0))] + [\frac{t}{2}\sin(2t)]_0^{2\pi}-\int_0^{2π}\frac{1}{2}\sin(2t)dt$
$\displaystyle [-2\pi^2(1) + 0] + [\frac{2\pi}{2}\sin(4\pi) - \frac{0}{2}\sin(0)] - [-\frac{1}{4}\cos(2t)]_0^{2\pi}$
$\displaystyle -2π^2 + 0 -[-\frac{1}{4}\cos(4\pi)-(-\frac{1}{4}\cos(0))]$
$\displaystyle -2π^2 - [-\frac{1}{4} + \frac{1}{4}]$
$-2\pi^2 - 0$
$-2\pi^2 ≈ -19.739208$
