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After solving my previous question, click here for question page, I tried to go up a notch and complicate the question just a bit further, turns out $\int e^x\sin(x)\cos(x)dx$ is much more different than $\int xe^x\sin(x)dx$. Much much much much much more. It's extremely longer (at least it was the way I went about it), after a long excruciating amount of time, I had gotten my final result, which needed simplifying. Here it is without simplying: $$ I = \frac{1}{2}e^x(\sin(x)+\cos(x))-\frac{1}{2}(I + \frac{1}{2}e^x(x+\frac{1}{2}\sin(2x))-\frac{1}{2}(xe^x-e^x)+\frac{1}{20}e^x(\sin(2x)-2\cos(2x))) + C $$

and here is what I got after attempting to simplify it, keep in mind I'm not even sure if this is correct. $$ 20I =e^x(10(\sin(x)+\cos(x)-x+\frac{5}{4}(\sin(2x) -x)) +\sin(2x)-2\cos(x)) - \frac{I}{2} +C$$

I really hope I didn't mess it up too much, if I've made a mistake in any of the things I've posted, please call me out on it. I didn't know what else to do after I tried to simplfy it, I know this is a lot to ask, and I'm sure you have better things to do, but just if by chance of a miracle, you feel like simplifying an extremely long equation, please, it would be really appreciated, and if not, then thank you for reading this far anyway. Thanks in advance. (Sorry for babbling on)

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Sam Chahine
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  • Without seeing all of your work, something does not look correct, see: http://www.wolframalpha.com/input/?i=integrate+e%5Ex+cos%28x%29+sin%28x%29 – Amzoti Jan 14 '14 at 13:54
  • That's what I use to check my answers, I saw that, but I was just hoping somehow, mathemagically my answer would equal that of wolframalpha – Sam Chahine Jan 14 '14 at 13:56
  • Well, I was just looking at the $x$ terms in your expression and those do not cancel, regardless of everything else. – Amzoti Jan 14 '14 at 13:57
  • you're right, now i'm thinking the longer equation might be closer to the correct answer – Sam Chahine Jan 14 '14 at 13:59

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$\int e^xsin(x)cos(x)dx = \dfrac{1}{2}\int e^xsin(2x) dx$

Then use integration by parts. You can easily solve this math by this way . There is no need of simplification .

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