Why do these integration steps hold true? Can someone explain the first three steps of the solution to this integral to me? I have searched but not found a lot: 
$$\begin{align}\int e^{ax}\cos(bx)dx &= \frac{1}{a}\int \cos(bx)de^{ax} \\
&= \frac{1}{a}e^{ax}\cos(bx)+\frac{b}{a}∫e^{ax}\sin(bx)dx\\
& = \frac{1}{a}e^{ax}\cos(bx)+ \frac{b}{a^2}∫\sin(bx)de^{ax}.
\end{align}$$
I know the basics of integration but this doesn't seem familiar. Would also really appreciate a link to a good source where I could catch up integration rules if anyone has any.

The first step involves a shorthand way of your usual $u$-sub.
Let $u=e^{ax}$.  Then, $du=ae^{ax}\ dx$ and $e^{ax}\ dx = \dfrac{1}{a} du$.
The second step involves integration by parts.
Let $v=\cos(bx)$ and $du=du$.  Then, $dv=-b\sin(bx)\ dx$ and $u=u$.
\begin{align*}
& \int e^{ax} \cos(bx) dx \\
=\ & \dfrac{1}{a}\int \cos(bx) du \\
=\ & \dfrac{1}{a}\bigg(u\cos(bx)-\int -be^{ax}\sin(bx)\ dx\bigg) \\
=\ & \dfrac{1}{a}e^{ax}\cos(bx)+\dfrac{b}{a}\int e^{ax}\sin(bx)\ dx \\
=\ & \dfrac{1}{a}e^{ax}\cos(bx)+\dfrac{b}{a}\cdot\dfrac{1}{a}\int u\sin(bx)\ du \\
=\ & \cdots \textrm{ apply integration by parts here}
\end{align*}
Does anything else need explaining?
