I'm trying to figure out the integral of (x^3)(sin x). Any pointers?
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This one requires integration by parts three times. For start, set $\displaystyle u=x^3$ & $\displaystyle dv=\sin x\,dx.$
Originally Posted by taichu I'm trying to figure out the integral of (x^3)(sin x). Any pointers? are you allowed to use Integration by parts? i.e, $\displaystyle \int udv = uv - \int vdu$ if so, do it three times..
Haha, I just figured that out now, and came to post my solution. What I got was -x^2 cos x + 3x^2 sin x - 6x cos x + 6 sin x + C.
Originally Posted by taichu Haha, I just figured that out now, and came to post my solution. What I got was -x^2 cos x + 3x^2 sin x - 6x cos x + 6 sin x + C. that is incorrect, i believe check the answer here
Originally Posted by taichu Haha, I just figured that out now, and came to post my solution. What I got was -x^2 cos x + 3x^2 sin x - 6x cos x + 6 sin x + C. it should be $\displaystyle x^3 \cos x + 3x^2 \sin x - 6x \cos x + 6 \sin x + C$
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