How Do You Solve the Integral of x^3*sin(2x) dx Using Integration by Parts?

  • Thread starter Thread starter 1MileCrash
  • Start date Start date
  • Tags Tags
    Integral
Click For Summary

Homework Help Overview

The discussion revolves around the integral of x^3*sin(2x) dx, focusing on the method of integration by parts and the challenges faced in applying this technique effectively.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of table integration and integration by parts, noting the reduction of the power of x with each application. There are questions about the correctness of the approach and the resulting expressions.

Discussion Status

Some participants suggest that the original poster is on the right track, pointing out that the new integrals involve smaller powers of x. Others express concerns about the correctness of the final expression obtained after several integrations.

Contextual Notes

There is mention of potential equivalent answers and the need for clarification on the expected final result, indicating that participants are navigating through different interpretations of the integral's solution.

1MileCrash
Messages
1,338
Reaction score
41

Homework Statement



[itex]\int x^{3}sin(2x) dx[/itex]

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.

[itex]\int x^{3}sin(2x) dx[/itex]

[itex]-\frac{1}{2}x^{3}cos2x + \frac{1}{2}\int3x^{2}cos2xdx[/itex]

I can solve that integral with integration by parts or using that same formula, am I going down the correct path for this?
 
Physics news on Phys.org
1MileCrash said:

Homework Statement



[itex]\int x^{3}sin(2x) dx[/itex]

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.

[itex]\int x^{3}sin(2x) dx[/itex]

[itex]-\frac{1}{2}x^{3}cos2x + \frac{1}{2}\int3x^{2}cos2xdx[/itex]

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!
 
1MileCrash said:

Homework Statement



[itex]\int x^{3}sin(2x) dx[/itex]

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).
1MileCrash said:
[itex]\int x^{3}sin(2x) dx[/itex]

[itex]-\frac{1}{2}x^{3}cos2x + \frac{1}{2}\int3x^{2}cos2xdx[/itex]

I can solve that integral with integration by parts or using that same formula, am I going down the correct path for this?
 
Since this integral involves [itex]x^3[/itex] and integration by parts (taking [itex]u= x^n[/itex] 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.
 
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.
 
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.
 

Similar threads

  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 105 ·
4
Replies
105
Views
11K
  • · Replies 22 ·
Replies
22
Views
3K
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
19
Views
2K
  • · Replies 23 ·
Replies
23
Views
2K