Answer
$\frac{3^{x}\sin(x)+\ln 3\cdot 3^{x}\cos{(x)}}{1+(\ln 3)^{2}}+C$
Work Step by Step
$$\int (3^{x}\cos(x))~dx$$
The integrand is a product, so we try writing $3^{x}\cos(x)dx=udv$ with:
$$u=3^{x},~~~~~~~~~~dv=\cos(x)dx$$
$$du=\frac{du}{dx}dx=\ln 3 \cdot 3^{x}dx,~~~~~~~~~~v=\sin(x)$$
By the Integration by Parts formula,
$$\int (3^{x}\cos(x))~dx=3^{x}\sin(x)-\int \sin(x) \ln 3 \cdot 3^{x}dx+C$$
$$\int (3^{x}\cos(x))~dx=3^{x}\sin(x)-\ln 3 \cdot\int 3^{x}\sin(x)dx+C$$
Using the same method, it follows:
$$\int 3^{x}\sin(x)dx=-3^{x}\cos{(x)}-\int -\cos(x) \ln 3 \cdot 3^{x}dx +C $$
$$\int 3^{x}\sin(x)dx=-3^{x}\cos{(x)}+\int \cos(x) \ln 3 \cdot 3^{x}dx+C $$
$$\int 3^{x}\sin(x)dx=-3^{x}\cos{(x)}+\ln 3 \cdot\int \cos(x) 3^{x}dx +C $$
so:
$$\int (3^{x}\cos(x))~dx=3^{x}\sin(x)-\ln 3 \cdot\int 3^{x}\sin(x)dx+C$$
$$\int (3^{x}\cos(x))~dx=3^{x}\sin(x)-\ln 3 \cdot(-3^{x}\cos{(x)}+\ln 3 \cdot\int \cos(x) 3^{x}dx)+C$$
$$\int (3^{x}\cos(x))~dx=3^{x}\sin(x)+\ln 3\cdot 3^{x}\cos{(x)}-(\ln 3)^{2}\cdot\int \cos(x) 3^{x}dx+C$$
$$I=3^{x}\sin(x)+\ln 3\cdot 3^{x}\cos{(x)}-(\ln 3)^{2}\cdot I+C$$
$$I+(\ln 3)^{2}\cdot I=3^{x}\sin(x)+\ln 3\cdot 3^{x}\cos{(x)}+C$$
$$I(1+(\ln 3)^{2})=3^{x}\sin(x)+\ln 3\cdot 3^{x}\cos{(x)}+C$$
$$I=\frac{3^{x}\sin(x)+\ln 3\cdot 3^{x}\cos{(x)}}{1+(\ln 3)^{2}}+C$$
$$\int 3^{x}\cos(x)~dx=\frac{3^{x}\sin(x)+\ln 3\cdot 3^{x}\cos{(x)}}{1+(\ln 3)^{2}}+C$$
