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rocomath
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#9
Jan18-08, 10:45 PM
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P: 1,757
Ok let's start from scratch.

[tex]I=\int x^2\cos{mx}dx[/tex]

[tex]u=x^2[/tex]
[tex]du=2xdx[/tex]

[tex]dV=\cos{mx}dx[/tex]
[tex]V=\frac{1}{m}\sin{mx}[/tex]

[tex]I=\frac{x^2}{m}\sin{mx}-\frac{2}{m}\int x\sin{mx}dx[/tex]

Now we have to do Parts again.

[tex]u=x[/tex]
[tex]du=dx[/tex]

[tex]dV=\sin{mx}dx[/tex]
[tex]V=\frac{-1}{m}\cos{mx}[/tex]

[tex]I=\frac{x^2}{m}\sin{mx}-\frac{2}{m}\left(\frac{-x}{m}\cos{mx}+\frac{1}{m}\int\cos{mx}dx\right)[/tex]

Now you can easily evaluate this Integral!!!