Finding the imaginary part of ∫e^(x(2+3i)) dx

[dcl]

Here is the problem:
$${\mathop{\rm Im}\nolimits} \int {e^{x(2 + 3i)} } dx$$

One sec, I'm having another go at it.
$$= {\mathop{\rm Im}\nolimits} \int {e^2 } e^{3ix} dx$$
$$= {\mathop{\rm Im}\nolimits} \int {e^2 } [\cos (3x) + i\sin (3x)]dx$$
$$\begin{array}{l}<br /> = \frac{{ - e^2 \cos (3t)}}{3} \\ <br /> \end{array}$$

How'd I go?

 

[arildno]

Not very well, I'm afraid..
Let us first consider the problem to calculate the antiderivative of the complex exponential,
$$\int{e}^{(2+3i)x)}dx$$

This is simply:
$$\int{e}^{(2+3i)x)}dx=\frac{1}{2+3i}{e}^{(2+3i)x)}+C$$

where C is an arbitrary complex constant (I'll set it in the following to 0, for simplicity)

We are to find the imaginary part:
$$Im(\frac{1}{2+3i}{e}^{(2+3i)x)})=Im(\frac{2-3i}{13}{e}^{(2+3i)x)})$$
or:
$$Im(\frac{1}{2+3i}{e}^{(2+3i)x)})=\frac{e^{2x}}{13}(2\sin(3x)-3\cos(3x))$$

 

[Zurtex]

$$e^{x(2+3i)} = e^{2x}e^{3xi} \neq e^{2}e^{3xi}$$

It's easier than that. Remember that:

$$\int e^{ax} dx = \frac{1}{a}e^{ax} + C$$

Once you integrate you then need to separate real from imaginary.

 

[dcl]

Stuffed up in the first step :(
Thanks guys. :)
prolly should goto bed now. :(