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

2 Answers
May 2, 2018

#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.

May 2, 2018

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))#