Evaluate the following integrals:

Let $I=\int e^{2 x} \cos (3 x+4) d x$

Integrating by parts

$I=e^{2 x} \frac{\sin (3 x+4)}{3}-\int 2 e^{2 x} \frac{\sin (3 x+4)}{3} d x$

$=\frac{1}{3} e^{2 x} \sin (3 x+4)-\frac{2}{3} \int e^{2 x} \sin (3 x+4) d x$

$=\frac{1}{3} e^{2 x} \sin (3 x+4)-\frac{2}{3}\left\{-e^{2 x} \frac{\cos (3 x+4)}{3}+\int 2 e^{2 x} \frac{\cos (3 x+4)}{3} d x\right\}$

$I=\frac{e^{2 x}}{9}[2 \cos (3 x+4)+3 \sin (3 x+4)]+c$

Hence,

$I=\frac{e^{2 x}}{9}[2 \cos (3 x+4)+3 \sin (3 x+4)]+c$