# int e^(4x)cos(2x) dx Find the indefinite integral

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...

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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  .

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