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Evaluating the integral, correct?
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Jan18-08, 11:05 PM
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12
rocomath
Quote by
Zack88
[tex]I=\frac{x^2}{m}\sin{mx}-\frac{2}{m}\left(\frac{-x}{m}\cos{mx}+\int\cos{mx}dx\right)[/tex]
ok where did the last [tex]\int\cos{mx}dx\right)[/tex] come from
By doing Parts again to evaluate ...
[tex]\int x\sin{mx}dx[/tex]