Integration using Euler's formula: Difference between revisions

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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 {{math|''e''^''ix''}} and {{math|''e''^{−−''ix''}}}, 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.

==Euler's formula==

Euler's formula states that

:<math>e^{ix} = \cos x + i\,\sin x.</math>

Substituting &minus;{{math|−''x''}} for {{math|''x''}} gives the equation

:<math>e^{-ix} = \cos x - i\,\sin x.</math>

These two equations can be solved for the sine and cosine:

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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}frac14\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 {{math|''e''^2''ix''&nbsp; +&nbsp; ''e''^{&minus;2−2''ix''}&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}frac14\int \left( e^{2ix} + 2 + e^{-2ix} \right) dx

\,&=\, \frac{1}{4}frac14\left(\frac{e^{2ix}}{2i} + 2x - \frac{e^{-2ix}}{2i}\right)+C \\[6pt]

&=\, \frac{1}{4}frac14\left(2x + \sin 2x\right) +C.

\end{align}</math>

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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}frac18\int \left(e^{2ix} - 2 + e^{-2ix}\right)\left(e^{4ix}+e^{-4ix}\right) dx \\[6pt]

&=\, -\frac{1}{8}frac18\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 {{math|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}frac18\sin 4x - \frac{1}{8}frac18 \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 {{math|cos&nbsp; ''x''}} is the real part of {{math|''e''^''ix''}}, 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>

</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}frac12 \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]] {{math|''u''&nbsp; {{=&nbsp;}} ''e''^''ix''}}, the result is the integral of a [[rational function]]:

:<math>-\frac{-i}{2}\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.

[[Category:Integral calculus]]