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

[asdf1]

i'm trying to solve an equation, but I'm stuck on this step:
(integral sign) e^2x*e^x(3sin2x+2cos2x)dx =?

 

[inha]

use the exponential form for the trig functions and multiply through. you'll get a few easy integrals of the form
$$\int e^{(a+ib)x}dx$$

 

[arildno]

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).

 

[asdf1]

ok, thanks! i'll try that~

 

[asdf1]

come to think of it... is there a quicker way?

 

[arildno]

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.

 

[asdf1]

haha... thanks! :)