Answer
$$\int e^{2x}\cos3xdx=\frac{2}{13}e^{2x}\cos3x+\frac{3}{13}e^{2x}\sin3x+C$$
Work Step by Step
$$A=\int e^{2x}\cos3xdx$$
Set $u=\cos3x$ and $dv=e^{2x}dx$
Then we have $du=-3\sin 3xdx$ and $v=\frac{1}{2}e^{2x}$
Using the formula $\int udv= uv-\int vdu$:
$$A=\frac{1}{2}e^{2x}\cos3x-\int\frac{1}{2}e^{2x}\times(-3)\sin3xdx$$ $$A=\frac{1}{2}e^{2x}\cos3x+\frac{3}{2}\int e^{2x}\sin3xdx$$
Set $u=\sin3x$ and $dv=e^{2x}dx$
Then we have $du=3\cos 3xdx$ and $v=\frac{1}{2}e^{2x}$
Using the formula $\int udv= uv-\int vdu$:
$$A=\frac{1}{2}e^{2x}\cos3x+\frac{3}{2}\Big(\frac{1}{2}e^{2x}\sin3x-\frac{3}{2}\int e^{2x}\cos3xdx\Big)$$
We notice that above, $\int e^{2x}\cos3xdx$ is exactly the given integral to solve, meaning it equals $A$.
Therefore, $$A=\frac{1}{2}e^{2x}\cos3x+\frac{3}{2}\Big(\frac{1}{2}e^{2x}\sin3x-\frac{3}{2}A\Big)$$ $$A=\frac{1}{2}e^{2x}\cos3x+\frac{3}{4}e^{2x}\sin3x-\frac{9}{4}A$$ $$\frac{13}{4}A=\frac{1}{2}e^{2x}\cos3x+\frac{3}{4}e^{2x}\sin3x+C$$ $$A=\frac{2}{13}e^{2x}\cos3x+\frac{3}{13}e^{2x}\sin3x+C$$
