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 Quote by Zack88 $$I=\frac{x^2}{m}\sin{mx}-\frac{2}{m}\left(\frac{-x}{m}\cos{mx}+\int\cos{mx}dx\right)$$ ok where did the last $$\int\cos{mx}dx\right)$$ come from
By doing Parts again to evaluate ...

$$\int x\sin{mx}dx$$