Solve Equation: (Integral) e^2x*e^x(3sin2x+2cos2x)dx

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i'm trying to solve an equation, but I'm stuck on this step:
(integral sign) e^2x*e^x(3sin2x+2cos2x)dx =?
 
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use the exponential form for the trig functions and multiply through. you'll get a few easy integrals of the form
\int e^{(a+ib)x}dx
 
If you don't know about the complex exponential, here's a technique using integration by parts (I'll let you tailor it to your specific example):
Suppose you are to find an anti-derivative (i.e, indefinite integral) of the function f(x)=e^{x}\sin(x), that is, you are to find J, where J is given as:
J=\int{e}^{x}\sin(x)dx(1)
The right-hand side can now be rewritten as:
\int{e}^{x}\sin(x)dx=e^{x}\sin(x)-\int{e}^{x}\cos(x)dx=e^{x}\sin(x)-e^{x}\cos(x)-\int{e}^{x}\sin(x)dx=e^{x}\sin(x)-e^{x}\cos(x)-J(2)
where I have used integration by parts twice, along with (1).
Thus, we have:
J=e^{x}\sin(x)-e^{x}\cos(x)-J\to{J}=\frac{e^{x}\sin(x)-e^{x}\cos(x)}{2}
(I've not bothered with the constant of integration; this should also be included in the final expression).
 
ok, thanks! i'll try that~
 
come to think of it... is there a quicker way?
 
Sure; we've got the "abra-kadabra" formula, but it is only taught to 50 year old professor with proven gentle disposition because of the formula's potential for abuse.
 
haha... thanks! :)
 
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