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 P: 1,757 Ok let's start from scratch. $$I=\int x^2\cos{mx}dx$$ $$u=x^2$$ $$du=2xdx$$ $$dV=\cos{mx}dx$$ $$V=\frac{1}{m}\sin{mx}$$ $$I=\frac{x^2}{m}\sin{mx}-\frac{2}{m}\int x\sin{mx}dx$$ Now we have to do Parts again. $$u=x$$ $$du=dx$$ $$dV=\sin{mx}dx$$ $$V=\frac{-1}{m}\cos{mx}$$ $$I=\frac{x^2}{m}\sin{mx}-\frac{2}{m}\left(\frac{-x}{m}\cos{mx}+\frac{1}{m}\int\cos{mx}dx\right)$$ Now you can easily evaluate this Integral!!!