Question

Integration!
\[\int x^2\sin x dx\]

Answer 1

I am guessing this is a fairly run of the mill integration by parts problem. I'll proceed that way unless someone tells me differently.

Answer 2

Yup,by parts

Answer 3

\[\LARGE x^2 \int\limits \sin xdx - \int\limits( \frac{d}{dx}x^2 \int\limits \sin xdx )dx\]

Answer 4

You can solve now,by parts twice fairly simple.

Answer 5

I sort of see what you're doing but I think that's a bit too streamlined for my current level of (lack of) confidence with integration by parts.

But anyway, here's what I've got so far.
\[u = x^2 \\ du = 2x\]\[dv = \sin x \\ v = \cos x\]

\[x^2\cos x - 2 \int x\cos x dx\]

Answer 6

Wait. \(-\cos x\)
\[-x^2\cos x + 2\int x \cos x dx\]

Answer 7

yup correct,now do same for x cosx

Answer 8

\[u = x \\ du = dx\]\[dv = \cos x \\ v = \sin x\]
So that gives me
\[-x^2\cos x + 2x\sin x - \int \sin x dx\]
One more.
\[-x^2\cos x + 2x\sin x + \cos x + C\]
Huge success?