Integral problem

**taichu** · November 1st 2007, 05:13 AM

I'm trying to figure out the integral of (x^3)(sin x). Any pointers?

**Krizalid** · November 1st 2007, 05:20 AM

This one requires integration by parts three times.

For start, set $u=x^3$ & $dv=\sin x\,dx.$

**kalagota** · November 1st 2007, 05:21 AM

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,
$\int udv = uv - \int vdu$

if so, do it three times..

**taichu** · November 1st 2007, 05:25 AM

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.

**Jhevon** · November 1st 2007, 08:23 PM

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

**kalagota** · November 1st 2007, 09:48 PM

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
$x^3 \cos x + 3x^2 \sin x - 6x \cos x + 6 \sin x + C$

Thread: Integral problem