TITLE: Integration by parts: $\int e^{ax}\cos(bx)\,dx$
QUESTION [12 upvotes]: I need to evaluate the following function and then check my answer by taking the derivative:
$$\int e^{ax}\cos(bx)\,dx$$
where $a$ is any real number and $b$ is any positive real number.

I know that you set $u=\cos(bx)$  and $dv=e^{ax} dx$,
and the second time you need to integrate again you set $u=\sin(bx)$ and $dv=e^{ax}dx$ again.
It eventually simplifies down to
$$\int e^{ax}\cos(bx)dx = \frac{1}{a}e^{ax}\cos(bx) + \frac{b}{a}\left(\frac{1}{a}e^{ax}\sin(bx) - \frac{b}{a}\int e^{ax}\cos(bx)\,dx\right).$$
Now I know to move the integral on the left side to the right side so that I can just divide by the constant to solve.
Here is my problem:
I know I need to solve the right side to be: 
$$\frac{e^{ax}\left(a\cos(bx) + b\sin(bx)\right)}{a^2+b^2} + C.$$
To divide by the constant, I multiplied everything on the right side by 
$$\frac{a^2}{b^2+1}.$$
but this leads me to get $b^2 + 1$ on the bottom instead of $a^2 + b^2$.
I will show what I am doing in detail:
After setting the initial integral equal to:
$$\frac{1}{a}e^{ax}\cos(bx) + \frac{b}{a}\left(\frac{1}{a}e^{ax}\sin(bx) - \frac{b}{a}\int e^{ax}\cos(bx)\,dx\right) + C$$
I simplify:
$$\int e^{ax}\cos(bx)\,dx = \frac{a^2}{b^2+1}\left(\frac{1}{a}e^{ax}\cos(bx) + \frac{b}{a}\left(\frac{1}{a}e^{ax}\sin(bx)\right)\right) + C$$
If this is already wrong, can someone point be in the right direction?  
If I have not gone wrong yet, I can edit to show the rest of my work.

REPLY [5 votes]: Another solution to your original question can be using complex numbers.
$$\begin{align}
I&=\displaystyle\int e^{ax}\cos {bx}dx \\
&=\Re\left(\displaystyle\int e^{ax}(\cos {bx}+i\sin {bx})\right)dx\\
&=\Re\left(\displaystyle\int e^{(a+ib)x}dx\right)\\
&=\Re\left(\dfrac{e^{(a+ib)x}}{a+ib}\right)\\
&=\Re\left(\dfrac{e^{ax}(\cos {bx}+i\sin {bx})}{a+ib}\right)\\
&=\Re\left(\dfrac{e^{ax}}{a^2+b^2}(\cos {bx}++i\sin {bx})(a-ib)\right)+C\\
\therefore I&=\dfrac{e^{ax}}{a^2+b^2}(a\cos {bx}+b\sin {bx})+C
\end{align}$$