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

May 2, 2018

$\int \frac{{e}^{i x} - {e}^{- i x}}{2} \mathrm{dx} = - i \cos x + C$

#### Explanation:

This requires using somewhat obscure trig identities, results which ultimately comes from Euler's formula.

The identities are ${e}^{i x} = \cos x + i \sin x$ and ${e}^{- i x} = \cos x - i \sin x$.

This turns $\int \frac{{e}^{i x} - {e}^{- i x}}{2} \mathrm{dx}$ into $\int \frac{\left(\cos x + i \sin x\right) - \left(\cos x - i \sin x\right)}{2} = \int \frac{2 i \sin x}{2} = \int \left(i \sin x\right)$

The $i$ is a constant, so this yields $i \int \sin x \mathrm{dx} = - i \cos x + C$, where $C$ is our constant of integration.

May 2, 2018

See below

#### Explanation:

We can use the identity sinhx=1/2·(e^x-e^(-x)) where $\sinh x$ is the hyperbolic sin of x

in our case $\sinh \left(i x\right)$ we know that $\int \sinh x = \cosh x$

Then $\int \sinh i x \mathrm{dx} = \frac{1}{i} \cosh i x + C = - i \cosh i x + C$

Other way is

1/2[inte^(ix)dx-inte^(-ix)dx]=1/2·1/ie^(ix)+1/2·1/ie^(-ix)=1/icoshix=-icoshix+C

Note that coshx=1/2·(e^x+e^(-x))