How Do You Integrate (e^ax)cos^2(2bx)dx Correctly?

[Geronimo85]

I'm supposed to integrate the following expression, and supposedly there is a very simple way to do so. Maple comes up with something rediculous, so I'd appreciate any input. Sorry about the short hand, don't know how to make everything pretty on here:

Integral[(e^ax)cos^2(2bx)dx] where a and b are positive constants

So far all I've got is:

(e^ax)cos^2(2bx)= (e^ax)*[(e^(i*2*b*x) - e^(-i*2*b*x))/2]^2

because: cosx = (e^ix - e^-ix)/2

squaring inside the brackets gets me:

(e^ax)*2[(e^(-4*b^2*x^2)-e^(4*b^2*x^2))/4]

I'm stuck and need to get this done for tomorrow

 

[HallsofIvy]

Why did you start a new thread for the same question? You are also still making the same mistake that was point out to you before:

Squaring inside the brackets does NOT give you

(e^ax)*2[(e^(-4*b^2*x^2)-e^(4*b^2*x^2))/4]

Because $(e^x)^2 = e^{2x}$, NOT $e^{x^2}$.

And, although I shouldn't have to say it if you are taking calculus, (a+ b)² is NOT a²+ b²!