Integration using Euler's formula: Difference between revisions

Revision as of 03:42, 16 November 2009

Revision as of 04:35, 23 November 2009

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:<math>\begin{align}

\int \sin^2 x \cos 4x \, dx \,

&=\, \int \left(\frac{e^{ix}+-e^{-ix}}{2i}\right)^2\left(\frac{e^{4ix}+e^{-4ix}}{2}\right) dx \\[6pt]

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

&=\, -\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 cos&nbsp;6''x''&nbsp;+-&nbsp;2&nbsp;cos&nbsp;4''x''&nbsp;+&nbsp;cos&nbsp;2''x'' and continue from there.

Either method gives

:<math>\int \sin^2 x \cos 4x \, dx \,=\, -\frac{1}{24}\sin 6x -+ \frac{1}{8}\sin 4x - \frac{1}{8}\sin 2x + C.</math>

==Using real parts==