# Integration of nasty trig product

• Dec 5th 2009, 12:49 PM
DangerousDave
Integration of nasty trig product
Hi all,

Would anybody be able to explain the technique for integrating something like this:

4p^2sin(10p^3-5) dp

I've been trying to do it by parts but with no success at all.
Any help appreciated.
• Dec 5th 2009, 12:54 PM
mr fantastic
Quote:

Originally Posted by DangerousDave
Hi all,

Would anybody be able to explain the technique for integrating something like this:

4p^2sin(10p^3-5) dp

I've been trying to do it by parts but with no success at all.
Any help appreciated.

Don't use integration by parts. Use the substitution $u = 10p^3 - 5$.
• Dec 5th 2009, 01:19 PM
DangerousDave
Cheers for response - but it still seems to be turning out pretty hideous. Am I to reformulate 4p^2 in terms of u too (and of course get du in terms of dp) to get a full expression in terms of u to integrate with respect to u? It seems like that results in a simlarly nasty-looking product in need of integration.
• Dec 5th 2009, 01:30 PM
mr fantastic
Quote:

Originally Posted by DangerousDave
Cheers for response - but it still seems to be turning out pretty hideous. Am I to reformulate 4p^2 in terms of u too (and of course get du in terms of dp) to get a full expression in terms of u to integrate with respect to u? It seems like that results in a simlarly nasty-looking product in need of integration.

$\int 4 p^2 \sin (10 p^3 - 5) \, dp$ becomes $\frac{2}{15} \int \sin (u) \, du$.
• Dec 5th 2009, 01:41 PM
DangerousDave
I think now I'm beginning to discover exactly where one of the main missing chunks in my education in calculus lies. I think I understand what's going on here:

du/dp = 30p^2, therefore dp = du/30p^2

that cancels immediately with 4P^2 to produce 2/15, which can then be removed in front of the integral.

My weakness is that I never really got to grips with the significance of the d-terms, and what it was permissible to do with them. 'With respect to' is such a non-mathematical sounding idea, that I didn't really know what to do with 'with respect to u divided by 30p^2' - that seems like comparing apples and oranges to me, and hard to see as a seamless part of the mathematical expression. Oh well, thanks for the help anyway.
• Dec 5th 2009, 01:45 PM
mr fantastic
Quote:

Originally Posted by DangerousDave
I think now I'm beginning to discover exactly where one of the main missing chunks in my education in calculus lies. I think I understand what's going on here:

du/dp = 30p^2, therefore dp = du/30p^2

that cancels immediately with 4P^2 to produce 2/15, which can then be removed in front of the integral.

My weakness is that I never really got to grips with the significance of the d-terms, and what it was permissible to do with them. 'With respect to' is such a non-mathematical sounding idea, that I didn't really know what to do with 'with respect to u divided by 30p^2' - that seems like comparing apples and oranges to me, and hard to see as a seamless part of the mathematical expression. Oh well, thanks for the help anyway.

You seem to have the right idea now.