Find of $\int e^x \cos(2x) dx$ I did the following.
Using the LIATE rule:
$$\begin{align*}
u &=& \cos(2x)\\
u\prime &=& -2 \sin(2x)\\
v &=& e^x\\
v\prime&=&e^x 
\end{align*}$$
We get:
$$\int e^x \cos(2x)dx = e^x \cos(2x) +2 \int e^x \sin(2x)dx $$
Now we do the second part.
$$\begin{align*}
u &=& \sin(2x)\\
u\prime &=& 2 \cos(2x)\\
v &=& e^x\\
v\prime&=&e^x 
\end{align*}$$
We get:
$$\int e^x \sin(2x)dx = e^x \sin(2x) -2 \int e^x \cos(2x)dx$$
Putting it together we get:
$\begin{align*}
\int e^x \cos(2x)dx &=& e^x \cos(2x) +2 \int e^x \sin(2x)dx \\
&=&e^x \cos(2x) + 2[e^x \sin(2x) -2 \int e^x \cos(2x)dx]\\
&=&e^x \cos(2x) + 2e^x \sin(2x) -4 \int e^x \cos(2x)dx
\end{align*}$
$\begin{align*}
\int e^x \cos(2x)dx &=&e^x \cos(2x) + 2e^x \sin(2x) -4 \int e^x \cos(2x)dx\\
5\int e^x \cos(2x)dx &=&e^x(\cos(2x) + 2 \sin(2x))\\
\int e^x \cos(2x)dx &=&\frac{e^x( \cos(2x) + 2 \sin(2x))}{5}
\end{align*}$
I am not sure if this is right, if it is right, is there a better way of doing this.

There is no need to ask whether it is right. You can differentiate to check.  A little problem: the arbitrary constant of integration is missing,
It is good to get accustomed to the approach you took, it is useful to master it.  Here is an alternative. My guess is that the answer will look like $(A\cos 2x+B\sin 2x)e^x$. Differentiate. We get 
$$(A\cos 2x+B\sin 2x)e^x+(-2A\sin 2x+2B\cos 2x)e^x.$$
This will match our integrand if $-2A+B=0$ and $2B+A=1$. Solve. We get $A=\frac{1}{5}$ and $B=\frac{2}{5}$. So our answer is $(A\cos 2x+B\sin 2x)e^x+C$, with $A$ and $B$ as just calculated. 
