Integration using Euler's formula: Difference between revisions

Revision as of 18:27, 26 March 2009

Revision as of 00:51, 1 July 2009

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&=\, -\frac{1}{8}\int \left(e^{6ix} + 2e^{4ix} + e^{2ix} + e^{-2ix} + 2e^{-4ix} + e^{-6ix}\right) dx.

\end{align}</math>

At this point we can either integrate directly, or we can first change the integrand to cos2cos&nbsp;6''x''&nbsp;+&nbsp;24&nbsp;cos&nbsp;4''x''&nbsp;+&nbsp;cos2cos&nbsp;2''x'' and continue from there.

Either method gives

:<math>\int \sin^2 x \cos 4x \, dx \,=\, -\frac{1}{4824}\sin 6x - \frac{1}{168}\sin 4x - \frac{1}{168}\sin 2x + C.</math>

==Using real parts==