# Integral using Euler's formula.

1. Mar 28, 2012

### 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?

2. Mar 28, 2012

### NeroKid

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

3. Mar 28, 2012

### cragar

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

4. Mar 28, 2012

### NeroKid

sry typos , its sin3x

5. Mar 28, 2012

### cragar

how come one part is not cos(3x)

6. Mar 28, 2012

### NeroKid

another typos , sry =='

7. Mar 29, 2012

### 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)

8. Mar 29, 2012

### NeroKid

but then ur integral cant become e^i3x now , can it

9. Mar 29, 2012

### cragar

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

10. Mar 29, 2012

### NeroKid

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

11. Mar 30, 2012