What is int (e^(ix)-e^(-ix))/(2i)dx?

1 Answer
Nov 6, 2015

-cosx+C

Explanation:

Using Euler's formula:

e^(ix) = cosx+isinx
e^(-ix) = cosx-isinx

e^(ix)-e^(-ix) = 2isinx => sinx= (e^(ix)-e^(-ix))/(2i)

int (e^(ix)-e^(-ix))/(2i)dx = int sinxdx = -cosx+C