Integration using Euler's formula: Difference between revisions

Revision as of 00:51, 1 July 2009

Revision as of 00:53, 1 July 2009

Line 35:

&=\, -\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 2coscos&nbsp;6''x''&nbsp;+&nbsp;42&nbsp;cos&nbsp;4''x''&nbsp;+&nbsp;2coscos&nbsp;2''x'' and continue from there.

Either method gives

:<math>\int \sin^2 x \cos 4x \, dx \,=\, -\frac{1}{2448}\sin 6x - \frac{1}{816}\sin 4x - \frac{1}{816}\sin 2x + C.</math>

==Using real parts==