Question

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

Answer 1

`int (e^(ix) - e^(-ix))/2dx = -icosx + C`

Explanation:

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

The identities are `e^(ix) = cosx + isinx` and `e^(-ix) = cosx - isinx`.

This turns `int (e^(ix) - e^(-ix))/2 dx` into `int ((cosx + isinx) - (cosx - isinx))/2 = int (2isinx)/2 = int(isinx)`

The `i` is a constant, so this yields `i int sinxdx = -i cosx + C`, where `C` is our constant of integration.

Answer 2

See below

Explanation:

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

in our case `sinh(ix)` we know that `intsinhx=coshx`

Then `intsinhixdx=1/icoshix+C=-icoshix+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))`