Integral of Product of Exponential with Sine or Cosine

Theorem

 * $\displaystyle \int e^{a x} \sin bx \ \mathrm d x = \frac {e^{a x} \left({a \sin bx - b \cos bx}\right)} {a^2 + b^2} + K$


 * $\displaystyle \int e^{a x} \cos bx \ \mathrm d x = \frac {e^{a x} \left({a \cos bx + b \sin bx}\right)} {a^2 + b^2} + K$

Proof
The technique to be used here is Integration by Parts:
 * $\displaystyle \int u \ dv = u v - \int v \ \mathrm d u$

For clarity, the arbitrary constant will be ignored. It is taken as read that it can be taken into account at the end of the integration process.

Let us put $\displaystyle S = \int e^{a x} \sin bx \ \mathrm d x$ and $\displaystyle C = \int e^{a x} \cos bx \ \mathrm d x$.

First we tackle $\displaystyle \int e^{a x} \sin bx \ \mathrm d x$:

Let $u = \sin bx$ and $dv = e^{ax}$.

Then:

So:

Now we take a look at $\displaystyle \int e^{a x} \cos bx \ \mathrm d x$:

Let $u = \cos bx$ and $dv = e^{ax}$.

Then:

So:

From these two results:

we can eliminate either $S$ or $C$ and obtain the result required.

Hence the results as given.