Answer
$$\int e^{3x}cos4x\,dx=\frac{1}{25}e^{3x}(4sin4x+3cos4x)+C$$
Work Step by Step
$$\int e^{3x}cos4x\,dx=\frac{1}{4}e^{3x}sin4x+\frac{3}{16}e^{3x}cos4x-\frac{9}{16}\int e^{3x}cos4x\,dx$$
$$\frac{25}{16}\int e^{3x}cos4x\,dx=\frac{1}{16}e^{3x}(4sin4x+3cos4x)$$
$$\int e^{3x}cos4x\,dx=\frac{1}{25}e^{3x}(4sin4x+3cos4x)+C$$
