Question

Integral x^3 sin x dx

Answer 1

\[\int\limits_{}^{} x^3 sin x dx\]

Answer 2

Do it by parts.
\[\int udv=uv-\int vdu\]

Answer 3

where
\[u=x^3\\dv=\sin x dx\\ v=-\cos x\\ du=3x^2 \]You have to redo the process three times.

Answer 4

\[u=x^3\]\[du = 3x^2 dx\]\[v = -cosx\]\[dv = sinx dx\]

Answer 5

Perfect.

Answer 6

\[\int\limits_{}^{} x^3 sinx dx = -x^3cosx + \int\limits_{}^{}cosx 3x^2dx\]

Answer 7

Yes, as you see you must repeat the integratin by parts in
\[3\int x^2\cos x dx\]

Answer 8

I have to go, but you can check your results here,
http://www.wolframalpha.com/input/?i=Integrate%5Bx%5E3+Sin%5Bx%5D%5D

Answer 9

I hope it helps.

Answer 10

Thank you!

Answer 11

\[u = x^2\]\[du = 2x dx\]\[v = sinx\]\[dv = cosx dx\]

Answer 12

\[\int\limits_{}^{} x^2cosxdx = x^2sinx -2\int\limits_{}^{}xsinxdx\]

Answer 13

Now what?