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**gconfused** Can someone please check that my answer for this question is correct:

Question:

$\displaystyle

A_n = \frac{2}{3} \int^{3}_{0} \frac{1}{2}x sin(\frac{n \Pi}{3} x) dx

$

My working out:

$\displaystyle

A_n = \frac{1}{2} \int^{3}_{0} x sin(\frac{n \Pi}{3} x) dx $

$\displaystyle = \frac{1}{2} {[x \frac{3}{n \Pi} cos (\frac{n \Pi}{3} x)]^{3}_{0} +\frac{3}{n \Pi} \int^{3}_{0} cos (\frac{n \Pi}{3} x) dx}$

$\displaystyle = \frac{1}{2} {[\frac{-9}{n \Pi} cos (n \Pi) - 0] + \frac{3}{n \Pi} [\frac{3}{n \Pi} sin (\frac{n \Pi}{3} x)]^{3}_{0}} $

$\displaystyle = \frac{9}{2 n^2 \Pi^2}sin(n \Pi) - \frac{9}{2 n \Pi} cos (n \Pi) $

$\displaystyle = \frac{9}{2 n \Pi} (-1)^(n+1) - frac{9}{2 n \Pi} cos (n \Pi) $

for some reason i think this looks wrong, can someone please confirm with me? thank you