Thread: How Good Am I?
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Hootenanny
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Jan16-07, 06:27 AM
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Quote Quote by murshid_islam View Post
does the answer come out to be [tex]-x^2\cos^{2}x + 2x\sin x + 2\cos x + C[/tex]?
Quote Quote by Gib Z View Post
I did integration by parts a few times, I got [tex]I=\frac{\sin^2 x}{2} - x^2 \cos x[/tex] :D I think thats right.

Note: This ones for x^2 sin x dx.
Neither is correct, but murshid is closer than Gib.