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`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 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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