Integration by complex exponential, extra term in argument of trig function

Question

I am trying to understand how best to integrate the following form using the complex exponential:

$\int(e^{ax}cos(b+cx)dx$

where a, b, c are real numbers.

I'm familiar with a method using the real part (or imaginary part) of a complex number. I have used this for the form:

$\int(e^{ax}cos(cx)dx$

but I run into trouble using this process with the first form due to the extra term $b$, from which I cannot factor out $x$ when it comes to:

$Real [ \int(e^{ax+(b+cx)i})dx] $

I am not sure whether it is correct to keep the $e^{b}$ term separate, and integrate it in its own right, being as it represents a multiplier that is entirely real.

Answer 1

This is an alternative method for solving $\int e^{ax}\cos(b+cx)dx$ by complex numbers. As you know \begin{align} \int e^{ax}\cos(b+cx)dx &= \mathbf{Re}\int e^{ax+(b+cx)i}dx \\ &= \mathbf{Re}\,e^{bi}\dfrac{e^{(a+ci)x}}{a+ci} \\ &= \mathbf{Re}\,\dfrac{a-ci}{a^2+c^2}e^{ax}e^{(b+cx)i} \\ &= \dfrac{e^x}{a^2+c^2}\Big(a\cos(b+cx)+c\sin(b+cx)\Big) \end{align}

Answer 2

HINT: using two times Integration by parts we get $$\frac{e^{a x} (a \cos (b+c x)+c \sin (b+c x))}{a^2+c^2}$$