Difficult integral by sub., parts & table

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Homework Statement


\int xe^{2x^2} cos(3x^2) dx
This is the hardest integral I've attempted so far. I've come up with an answer that fits a table in my book but I'm not sure if I arrived there correctly. Thanks for reading!


Homework Equations


\int e^{au} cos\ bu\ du\ = \frac{e^{au}}{a^2+b^2}(a\ cos\ bu\ + b\ sin\ bu)+C


The Attempt at a Solution


http://www.mcp-server.com/~lush/shillmud/int2.7.JPG
http://www.mcp-server.com/~lush/shillmud/int2.72.JPG
 
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substitute:

u=x^2

du=2x dx

I=\int xe^{2x^2}\cos(3x^2){\rm dx}\,=\, {1\over 2} \int e^{2u}\cos(3u) {\rm du}

I have done some shortcut here, with use of the formula, your number two:

I={1\over 26} e^{2u} (2 \cos(3u)\,+\,3\sin(3u))\,+\,C

I={1\over 26} e^{2x^2} (2 \cos(3x^2)\,+\,3\sin(3x^2))\,+\,C

this is the same as integrator...

http://integrals.wolfram.com/

check it out
 
Here's the solution, I didn't simplify it any more like I should have, but I hope you understand the basic process. I checked my answer with CAS and it's correct, so don't worry about that.
http://img411.imageshack.us/img411/5615/file0001su6.jpg
 
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How do you submit scanned work onto here?
Thanks
 
SavvyAA3 said:
How do you submit scanned work onto here?
Thanks

Assuming the scanned image is in a format the forum recognises, jpeg,mpeg and so on either click on the icon in yellow with a black mountain or two and a black sun, visible in the new reply or go advanced: edit windows, not quick post. Or type [noparse]*url [/noparse] around the files url.

To get a url for an image stored on your hard drive you will need to upload it to a web host facility such as photobucket or freewebs. And then cut and paste the url. If not then use the paper clip to attach files directly from your hard drive to the forum. With this option there will be a delay while the forum scans the file for viruses, but any format that is listed there from jpeg to gif are acceptable.
 
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Thanks alot!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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