Need some help with an integral.

[1MileCrash]

Homework Statement

$\int x^{3}sin(2x) dx$

Relevant equations

The Attempt at a Solution

We are doing table integration - where I use a formula for integrals. I'm having trouble here because I keep getting new integrals through my work which don't seem to be leading anywhere.

$\int x^{3}sin(2x) dx$

$-\frac{1}{2}x^{3}cos2x + \frac{1}{2}\int3x^{2}cos2xdx$

I can solve that integral with integration by parts or using that same formula, am I going down the correct path for this?

 

[saxen]

Homework Statement

$\int x^{3}sin(2x) dx$

Relevant equations

The Attempt at a Solution

We are doing table integration - where I use a formula for integrals. I'm having trouble here because I keep getting new integrals through my work which don't seem to be leading anywhere.

$\int x^{3}sin(2x) dx$

$-\frac{1}{2}x^{3}cos2x + \frac{1}{2}\int3x^{2}cos2xdx$

I can solve that integral with integration by parts or using that same formula, am I going down the correct path for this?

This integral is solvable with integration by parts. Notice that you started with x^3 and after your first integration you have x^2. So just keep integrating!

 

[Mark44]

Homework Statement

$\int x^{3}sin(2x) dx$

Relevant equations

The Attempt at a Solution

We are doing table integration - where I use a formula for integrals. I'm having trouble here because I keep getting new integrals through my work which don't seem to be leading anywhere.

Actually, you are getting somewhere. Notice that the new integral has a smaller power of x (x²). At your next step, you should get that down to x, and then one more step will result in an integral that involves only sin(2x) or cos(2x).

$\int x^{3}sin(2x) dx$

$-\frac{1}{2}x^{3}cos2x + \frac{1}{2}\int3x^{2}cos2xdx$

I can solve that integral with integration by parts or using that same formula, am I going down the correct path for this?

 

[HallsofIvy]

Since this integral involves $x^3$ and integration by parts (taking $u= x^n$ each time) reduces the power of x by 1, you need to do three consecutive integrations by parts to get to a integral of a trig function only.

 

[1MileCrash]

Alright guys, I see what you're saying. However, while the process does eventually lead me to a final expression without integrals, it doesn't seem to be correct.

http://imageshack.us/photo/my-images/689/integral.png/

That's a link to the work I did.. I didn't think itexing all of that would be feasible. Hopefully it's legible.

 

[ArcanaNoir]

I got the same answer you did. Perhaps it is a case of equivalent answers, where your answer could be re-arranged to match the given answer. If you post what the answer should be I can check for equivalence on my calculator very easily.