how to integrate cos(2x)/(2e^x) ?

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)

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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 .