- #1

cragar

- 2,552

- 3

And then I use Eulers formula to substitute in for the sine terms.

So I have [itex] \int e^{2x}e^{ix}e^{2ix} [/itex]

then I combine everything so i get

[itex] e^{(2+3i)x} [/itex]

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?