Calculus of a Single Variable, Chapter 8, 8.2, Section 8.2, Problem 30

int e^(4x)cos(2x)dx
To solve, apply integration by parts int u dv = u*v - int vdu .
In the given integral, the let the u and dv be:

u = e^(4x)
dv = cos(2x)dx

Then, take the derivative of u to get du. Also, take the integral of dv to get v.

du = e^(4x)*4dx
du = 4e^(4x)dx
intdv = int cos(2x)dx
v = (sin(2x))/2

Substituting them to the integration by parts formula yields
int e^(4x)cos(2x)dx= e^(4x)*(sin(2x))/2 - int (sin(2x))/2 * 4e^(4x)dx
int e^(4x)cos(2x)dx= (e^(4x)sin(2x))/2 - int 2e^(4x)sin(2x)dx
For the integral at the right side, apply integration by parts again. Let the u and dv be:

u = 2e^(4x)
dv = sin(2x)dx

Take the derivative of u and take the integral of dv to get du and v, respectively.

du = 2e^(4x)*4dx
du = 8e^(4x)dx
int dv = int sin(2x)dx
v = -cos(2x)/2

Plug-in them to the formula of integration by parts.
int e^(4x)cos(2x)dx= (e^(4x)sin(2x))/2 - int 2e^(4x)sin(2x)dx
int e^(4x)cos(2x)dx= (e^(4x)sin(2x))/2 - [2e^(4x)*(-(cos(2x))/2) - int (-(cos(2x))/2)*8e^(4x)dx]
int e^(4x)cos(2x)dx= (e^(4x)sin(2x))/2 - [-e^(4x)cos(2x)+int 4e^(4x)cos(2x)dx]
int e^(4x)cos(2x)dx= (e^(4x)sin(2x))/2 +e^(4x)cos(2x) - 4int e^(4x)cos(2x)dx
Since the integrals at the left and right side of the equation are like terms, bring them together on one side.
int e^(4x)cos(2x)dx+4int e^(4x)cos(2x)dx= (e^(4x)sin(2x))/2 +e^(4x)cos(2x)
The left side simplifies to
5int e^(4x)cos(2x)dx= (e^(4x)sin(2x))/2 +e^(4x)cos(2x)
Isolating the integral, the equation becomes
int e^(4x)cos(2x)dx= ((e^(4x)sin(2x))/2 +e^(4x)cos(2x)) * 1/5
int e^(4x)cos(2x)dx= (e^(4x)sin(2x))/10 +(e^(4x)cos(2x))/5
Since it is an indefinite integral, add C at the right side.
Therefore,
int e^(4x)cos(2x)dx= (e^(4x)sin(2x))/10 +(e^(4x)cos(2x))/5+C .