Integral using Euler's formula.

[cragar]

If I have \int e^{2x}sin(x)sin(2x)
And then I use Eulers formula to substitute in for the sine terms.
So I have \int e^{2x}e^{ix}e^{2ix}
then I combine everything so i get
e^{(2+3i)x}
so then we integrate the exponential, so we divide by 2+3i
and then i multiply by the complex conjugate. now since sine is the imaginary part of his
formula I took the imaginary part when I back substituted for e^(3i)
but I didn't get the correct answer doing this, so am i not using Eulers formula correctly?

 

[NeroKid]

e^i3x = sin3x+isin2x , so the imaginary part is different from i(sinxsin2x)

 

[cragar]

why does e^i3x = sin3x+isin2x , i guess I am not seeing it off hand I probably should look at it more and try to manipulate it more.

 

[NeroKid]

sry typos , its sin3x

 

[cragar]

how come one part is not cos(3x)

 

[NeroKid]

another typos , sry =='

 

[cragar]

But we could get it in the form of
sin(x)e^{2ix}=isin(2x)sin(x)+cos(2x)sin(x)
Do we need to get an expression where we have just exponentials on the left hand side
and then isin(x)sin(2x)+cos(2x)cos(x)

 

[NeroKid]

but then ur integral can't become e^i3x now , can it

 

[cragar]

ok, I am not sure exactly what you mean, How do you recommend I approach the problem.

 

[NeroKid]

sina*sinb = -0.5[cos(a+b)+cos(a-b)] , then u have 2 solvable integrals

 

[cragar]

oh i see thanks for your answer.