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.
 A: 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?
A: As it was pointed out in one comment, excluding the second and the last equality this just uses "integration by parts". Now, for the second and last equality the normal Riemann integral was rewritten into something called a Riemann-Stiltjes integral. The basic idea here, is that in the definition of the Riemann integral one just uses the difference between points of an interval in the Riemann sum whereas for the definition of the Riemann-Stiltjes integral one can also use a certain "weight" as a function. Lets call it $f$ then instead of having an expression like $\int \dots dx$ one would write $\int \dots df(x)$ to indicate this integral.
A good first source for this would be wikipedia I guess link. 
Also notice that if this weighted function $f$ is actually continuously differentiable one has the equality:
$$
\int gf' dx = \int g df(x)
$$
and that is what precisely was done in equal sign two and the last one. Also note that integration by parts also holds for this Riemann-Stieltjes integral. 
A: $\int e^{ax}\cos(bx)dx = \frac{1}{a}\int \cos(bx)de^{ax} = \frac{1}{a}e^{ax}\cos(bx)+\frac{b}{a}∫e^{ax}\sin(bx)dx = \frac{1}{a}e^{ax}\cos(bx)+ \frac{b}{a^2}∫\sin(bx)de^{ax}$
This solution is basically an application of the method known as integration by parts. So, are you familiar with this method?
If you are then perhaps what is confusing you is just the notation. The expression 
$$\int e^{ax}\cos(bx)dx$$
is the integral of a product of the two functions, $e^{ax}$ and $\cos(bx)$. Forget about the written down solution and see if you can apply integration by parts to this product. You will need to integrate the first of them and differentiate the second.
I hope that helps!
