Integration using Euler's formula: Difference between revisions

Revision as of 15:59, 18 July 2020

Revision as of 02:31, 23 January 2021

Line 64:

Using Euler's identity, this integral becomes

:<math>\frac12 \int \frac{6 + e^{2ix} + e^{-2ix} }{e^{ix} + e^{-ix} + e^{3ix} + e^{-3ix}} \, dx.</math>

If we now make the [[integration by substitution|substitution]] {{<math|''>u'' {{=}} ''e''<sup>''^{ix''}</supmath>}}, the result is the integral of a [[rational function]]:

:<math>-\frac{i}{2}\int \frac{1+6u^2 + u^4}{1 + u^2 + u^4 + u^6}\,du.</math>

One may proceed using [[partial fraction decomposition]].