Write `int (cos2x)/(2e^x) dx = int (cos2x)/2e^(-x) dx`

Using the formula `cos2x = (e^(2ix) + e^(-2ix))/2` (ref Euler's formula)

we have

`int (cos2x)/(2e^x) dx = int (e^(2ix) + e^(-2ix))/4e^(-x) dx`

Multiplying out

`int (cos2x)/(2e^x) dx = 1/4(int e^((2i-1)x) dx + int e^(-(2i+1)x) dx)`

`= 1/4( 1/(2i-1)e^((2i-1)x)-1/(2i+1)e^(-(2i+1)x)) + C`

`= (e^(-x)/4)((2i+1)e^(2ix) - (2i-1)e^(-2ix))/((2i-1)(2i+1)) + C`

`= (e^(-x)/(4(4i^2-1)))((e^(2ix) + e^(-2ix)) +2i(e^(2ix)-e^(-2ix))) + C`

Using the above formula for `cos2x` and `sin2x = (e^(2ix) -e^(-2ix))/(2i)`

`int (cos2x)/(2e^x) dx = -e^(-x)/20(2cos2x + 4i^2sin2x) + C`

`= e^(-x)/10(2sin2x -cos2x) + C`

answer (using Euler's formula as an alternative to integration by parts)

`int (cos(2x))/(2e^x) dx`

`= int (cos(2x) e^(-x))/2 dx`

To integrate, use integration by parts. The formula is `int u dv = uv - int vdu` .

So let,

`u = (cos(2x))/2`                     and             `dv= e^(-x)`

Then, determine du and v.

`du=1/2(-sin(2x))*2 dx`                         `v = int e^(-x) dx`

`du=-sin(2x) dx`                                 `v=-e^(-x)`

Substitute u, v and du to the formula of integration by parts.

`int (cos(2x)e^(-x))/2dx = (cos(2x))/2*(-e^(-x)) - int (-e^(-x))(-sin(2x)dx)`

                           `= -(cos(2x)e^(-x))/2 - int sin(2x)e^(-x)dx`

To evaluate `int sin(2x)e^(-x) dx` , use integration by parts again.So let,

`u = sin(2x)`                  and               `dv = e^(-x)dx`

Solve for du and v.

`du=cos(2x)*2dx`                              `v =int e^(-x)dx`  

`du=2cos(2x)dx `                                 `v=-e^(-x)`

Substitute u, v and du to the formula of integration by parts.

`int (cos(2x)e^(-x))/2dx = -(cos(2x)e^(-x))/2 - [-sin(2x)e^(-x)+2int cos(2x)e^(-x)dx]`

                         `= -(cos(2x)e^(-x))/2 +sin(2x)e^(-x) -2intcos(2x)e^(-x)dx`

To have the same integrand, express `2int cos(2x)e^(-x) dx` as `4 int (cos(2x)e^(-x))/2 dx` .

`int (cos (2x)e^(-x))/2dx= -(cos(2x)e^(-x))/2 +sin(2x)e^(-x) -4int(cos(2x)e^(-x))/2dx`

Now that both sides have the same integrand `(cos(2x)e^(-x))/2` , bring together the integrals. So, add both sides by `4 int (cos(2x) e^(-x))/2dx` .

`int (cos (2x)e^(-x))/2dx+4int (cos(2x)e^(-x))/2 dx=-(cos(2x)e^(-x))/2 +sin(2x)e^(-x)`

`5int (cos(2x)e^(-x))/2 dx =-(cos(2x)e^(-x))/2 +sin(2x)e^(-x)`

To isolate `int (cos(2x)e^(-x))/2 dx` , divide both sides by 5.

`int (cos (2x)e^(-x))/2dx = [-(cos(2x)e^(-x))/2 +sin(2x)e^(-x)]/5`

                           `= -(cos(2x)e^(-x))/10 + ( sin(2x)e^(-x))/5`

                           `= (sin(2x)e^(-x))/5 - (cos(2x)e^(-x))/10`

Since we have an indefinite integral, then

`int (cos(2x))/(2e^x) dx=(sin(2x)e^(-x))/5 - (cos(2x)e^(-x))/10+C` .