Integration using Euler's formula: Difference between revisions

Revision as of 19:42, 28 January 2012

Revision as of 20:53, 30 January 2012

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:<math>\frac{1}{2}\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]] ''u''&nbsp;=&nbsp;''e''^''ix'', the result is the integral of a [[rational function]]:

:<math>\frac{1}{22i}\int \frac{6 1+ u6u^2 + u^{-2}4}{u1 + u^{-1}2 + u^34 + u^{-3}6}\,\frac{du}{iu}.</math>

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

Any rational function is integrable (using, for example, [[partial fractions in integration|partial fractions]]), and therefore any rational expression involving trigonometric functions may be integrated as well.