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In [[integral calculus]], [[complex number]]s and [[Euler's formula]] may be used to evaluate [[integral]]s involving [[trigonometric functions]]. Using Euler's formula, any trigonometric function may be written in terms of ''e''<sup>''ix''</sup> and ''e''<sup>&minus;''ix''</sup>, and then integrated. This technique is often simpler and faster than using [[trigonometric identities]] or [[integration by parts]], and is sufficiently powerful to integrate any rational expression involving trigonometric functions.
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==Euler's formula==
Euler's formula states that
:<math>e^{ix} = \cos x + i\,\sin x.</math>
Substituting &minus;''x'' for ''x'' gives the equation
:<math>e^{-ix} = \cos x - i\,\sin x.</math>
These two equations can be solved for the sine and cosine:
:<math>\cos x = \frac{e^{ix} + e^{-ix}}{2}\quad\text{and}\quad\sin x = \frac{e^{ix}-e^{-ix}}{2i}.</math>
 
==Simple example==
Consider the integral
:<math>\int \cos^2 x \, dx.</math>
The standard approach to this integral is to use a [[half-angle formula]] to simplify the integrand.  We shall use Euler's identity instead:
:<math>\begin{align}
\int \cos^2 x \, dx \,&=\, \int \left(\frac{e^{ix}+e^{-ix}}{2}\right)^2 dx \\[6pt]
&=\, \frac{1}{4}\int \left( e^{2ix} + 2 + e^{-2ix} \right) dx
\end{align}</math>
At this point, it would be possible to change back to real numbers using the formula ''e''<sup>2''ix''</sup>&nbsp;+&nbsp;''e''<sup>&minus;2''ix''</sup>&nbsp;=&nbsp;2&nbsp;cos&nbsp;2''x''. Alternatively, we can integrate the complex exponentials and not change back to trigonometric functions until the end:
:<math>\begin{align}
\frac{1}{4}\int \left( e^{2ix} + 2 + e^{-2ix} \right) dx
\,&=\, \frac{1}{4}\left(\frac{e^{2ix}}{2i} + 2x - \frac{e^{-2ix}}{2i}\right)+C \\[6pt]
&=\, \frac{1}{4}\left(2x + \sin 2x\right) +C.
\end{align}</math>
 
==Second example==
Consider the integral
:<math>\int \sin^2 x \cos 4x \, dx.</math>
This integral would be extremely tedious to solve using trigonometric identities, but using Euler's identity makes it relatively painless:
:<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==
In addition to Euler's identity, it can be helpful to make judicious use of the [[real part]]s of complex expressions. For example, consider the integral
:<math>\int e^x \cos x \, dx.</math>
Since cos&nbsp;''x'' is the real part of ''e''<sup>''ix''</sup>, we know that
:<math>\int e^x \cos x \, dx \,=\, \operatorname{Re}\int e^x e^{ix}\, dx.</math>
The integral on the right is easy to evaluate:
:<math>\int e^x e^{ix} \, dx \,=\, \int e^{(1+i)x}\,dx \,=\, \frac{e^{(1+i)x}}{1+i} + C.</math>
Thus:
:<math>\begin{align}
\int e^x \cos x \, dx \,&=\, \operatorname{Re}\left\{\frac{e^{(1+i)x}}{1+i}\right\} + C \\[6pt]
&=\, e^x\operatorname{Re}\left\{\frac{e^{ix}}{1+i}\right\} +C \\[6pt]
&=\, e^x\operatorname{Re}\left\{\frac{e^{ix}(1-i)}{2}\right\} +C \\[6pt]
&=\, e^x\,\frac{\cos x + \sin x}{2} +C.
\end{align}
</math>
 
==Fractions==
In general, this technique may be used to evaluate any fractions involving trigonometric functions. For example, consider the integral
:<math>\int \frac{1+\cos^2 x}{\cos x + \cos 3x} \, dx.</math>
Using Euler's identity, this integral becomes
:<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''<sup>''ix''</sup>, the result is the integral of a [[rational function]]:
:<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 fraction involving trigonometric functions may be integrated as well.
 
==External links==
*[http://www.docstoc.com/docs/159557614/Exponential-Circular-Integrals/ Evaluation of difficult integrals using complex numbers and Euler's formula]
[[Category:Integral calculus]]

Latest revision as of 21:54, 20 September 2014

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