`int e^(-theta) cos(2 theta) d theta` Evaluate the integral

2∫(-2excos(2x))dx+2exsin(2x)-1excos(2x)=2(-2∫1excos(2x)dx)+2exsin(2x)-1excos(2x)

Use integration by parts:

`int (u)dv = uv-intvdu`

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

Then,

`du = -2sin2xdx`

`v = int(e^-x)dx = -1/e^x`

Thus,

`int (1/e^xcos(2x))dx =(cos(2x)⋅(-1/e^x)−int((-1/e^x)⋅(-2sin(2x))dx))=(-int(2/e^xsin(2x)dx-1/e^xcos(2x)))`

Use the constant multiple rule

`-int(2/e^xsin(2x)dx-1/e^xcos(2x))=-(2int(1/e^xsin(2x)dx)-1/e^xcos(2x))`

Use integration by parts again for the first part

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

Then,

`du = 2cos(2x)dx and v = int(e^-x)dx = -1/e^-x`

Integral becomes

`-2int(1/e^xsin(2x)dx-1/e^xcos(2x))=-2(sin(2x)(-1/e^x)-int((-1/e^x)2cos(2x)dx))`

`-1/e^xcos(2x)=-2(-int((-2/e^xcos(2x))dx-1/e^xsin(2x)))-1/e^xcos(2x)`

Apply constant multiple rule

`2int((-2/e^xcos(2x))dx)+2/e^xsin(2x)-1/e^xcos(2x)=2(-2int(1/e^xcos(2x)dx))+2/e^xsin(2x)-1/e^xcos(2x)`

Simplify

`int(1/e^xcos(2x))dx=1/(5e^x)(2sin(2x)-cos(2x))+C`