# Thread: Integration of nasty trig product

1. ## 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.

2. 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 $u = 10p^3 - 5$.

3. 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.

4. 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 $\frac{2}{15} \int \sin (u) \, du$.

5. 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.

6. 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.