How do you find the antiderivative of (cos 2x)e^(cos 2x)(cos2x)ecos2x?

1 Answer
Mar 6, 2015

I don't think you can (in finite form).

You can find the antiderivative of (sin2x)e^cos(2x)(sin2x)ecos(2x) or of (cos2x)e^sin(2x)(cos2x)esin(2x) by substitution.

The antiderivative of (sin2x)e^cos(2x)(sin2x)ecos(2x) is -1/2e^cos(2x)+C 12ecos(2x)+C.
(Let u=cos(2x)u=cos(2x) , so du=-2sin(2x)dxdu=2sin(2x)dx)

The antiderivative of (cos2x)e^sin(2x)(cos2x)esin(2x) is 1/2e^sin(2x)+C 12esin(2x)+C.
(Let u=sin(2x)u=sin(2x), so du=2cos(2x)dxdu=2cos(2x)dx)