# Evaluate the integral: integral of 3*e^(2x) cos(2x + 1) dx.

## Question:

Evaluate the integral: {eq}\; \int 3 e^{2x} \cos(2x + 1) \, \mathrm{d}x {/eq}.

## Integration:

To evaluate the integral integration by parts will be used as it is the product of two functions. After solving the integral integration constant will be used. Differentiation is the reverse process of integration

## Answer and Explanation:

{eq}I=\int 3e^{2x}\cos(2x+1)dx\\ =\frac{3e^{2x}\sin(2x+1)}{2}-\int 3e^{2x}\sin(2x+1)dx\\ =\frac{3e^{2x}\sin(2x+1)}{2}-3\left [ \frac{-e^{2x}\cos(2x+1)}{2}+\int e^{2x}\cos(2x+1)dx \right ]\\ I=\frac{3e^{2x}\sin(2x+1)}{2}+\frac{3e^{2x}\cos(2x+1)}{2}-\int 3e^{2x}\cos(2x+1)dx\\ 4I=\frac{3e^{2x}\sin(2x+1)}{2}+\frac{3e^{2x}\cos(2x+1)}{2}\\ I=\frac{3e^{2x}\sin(2x+1)}{8}+\frac{3e^{2x}\cos(2x+1)}{8} {/eq}