Integral of Product of Exponential with Sine or Cosine

Theorem

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


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

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

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 $$S = \int e^{a x} \sin bx \, dx$$ and $$C = \int e^{a x} \cos bx \, dx$$.

First we tackle $$\int e^{a x} \sin bx \, dx$$:

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

Then:

$$ $$

So:

$$ $$ $$

Now we take a look at $$\int e^{a x} \cos bx \, dx$$:

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.