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

• asdf1
In summary, the conversation is about solving an integral equation involving exponential and trigonometric functions. The recommended method is to use the exponential form for the trigonometric functions and integration by parts. Another option is to use the "abra-kadabra" formula, but it is only taught to experienced professors due to its potential for misuse.

#### asdf1

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

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! :)