Integration using Euler's formula

In integral calculus, complex numbers and Euler's formula may be used to evaluate integrals involving trigonometric functions. Using Euler's formula, any trigonometric function may be written in terms of e^ix and 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

$e^{ix}=\cos x+i\,\sin x.$

Substituting −x for x gives the equation

$e^{-ix}=\cos x-i\,\sin x.$

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

$\cos x={\frac {e^{ix}+e^{-ix}}{2}}\quad {\text{and}}\quad \sin x={\frac {e^{ix}-e^{-ix}}{2i}}.$

Simple example

Consider the integral

$\int \cos ^{2}x\,dx.$

The standard approach to this integral is to use a half-angle formula to simplify the integrand. We shall use Euler's identity instead:

${\begin{aligned}\int \cos ^{2}x\,dx\,&=\,\int \left({\frac {e^{ix}+e^{-ix}}{2}}\right)^{2}dx\\[6pt]&=\,{\frac {1}{4}}\int \left(e^{2ix}+2+e^{-2ix}\right)dx\end{aligned}}$

At this point, it would be possible to change back to real numbers using the formula e^2ix + e^−2ix = 2 cos 2x. Alternatively, we can integrate the complex exponentials and not change back to trigonometric functions until the end:

${\begin{aligned}{\frac {1}{4}}\int \left(e^{2ix}+2+e^{-2ix}\right)dx\,&=\,{\frac {1}{4}}\left({\frac {e^{2ix}}{2i}}+2x-{\frac {e^{-2ix}}{2i}}\right)+C\\[6pt]&=\,{\frac {1}{4}}\left(2x+\sin 2x\right)+C.\end{aligned}}$

Second example

Consider the integral

$\int \sin ^{2}x\cos 4x\,dx.$

This integral would be extremely tedious to solve using trigonometric identities, but using Euler's identity makes it relatively painless:

${\begin{aligned}\int \sin ^{2}x\cos 4x\,dx\,&=\,\int \left({\frac {e^{ix}-e^{-ix}}{2i}}\right)^{2}\left({\frac {e^{4ix}+e^{-4ix}}{2}}\right)dx\\[6pt]&=\,-{\frac {1}{8}}\int \left(e^{2ix}-2+e^{-2ix}\right)\left(e^{4ix}+e^{-4ix}\right)dx\\[6pt]&=\,-{\frac {1}{8}}\int \left(e^{6ix}-2e^{4ix}+e^{2ix}+e^{-2ix}-2e^{-4ix}+e^{-6ix}\right)dx.\end{aligned}}$

At this point we can either integrate directly, or we can first change the integrand to cos 6x - 2 cos 4x + cos 2x and continue from there. Either method gives

$\int \sin ^{2}x\cos 4x\,dx\,=\,-{\frac {1}{24}}\sin 6x+{\frac {1}{8}}\sin 4x-{\frac {1}{8}}\sin 2x+C.$

Using real parts

In addition to Euler's identity, it can be helpful to make judicious use of the real parts of complex expressions. For example, consider the integral

$\int e^{x}\cos x\,dx.$

Since cos x is the real part of e^ix, we know that

$\int e^{x}\cos x\,dx\,=\,\operatorname {Re} \int e^{x}e^{ix}\,dx.$

The integral on the right is easy to evaluate:

$\int e^{x}e^{ix}\,dx\,=\,\int e^{(1+i)x}\,dx\,=\,{\frac {e^{(1+i)x}}{1+i}}+C.$

Thus:

${\begin{aligned}\int e^{x}\cos x\,dx\,&=\,\operatorname {Re} \left\{{\frac {e^{(1+i)x}}{1+i}}\right\}+C\\[6pt]&=\,e^{x}\operatorname {Re} \left\{{\frac {e^{ix}}{1+i}}\right\}+C\\[6pt]&=\,e^{x}\operatorname {Re} \left\{{\frac {e^{ix}(1-i)}{2}}\right\}+C\\[6pt]&=\,e^{x}\,{\frac {\cos x+\sin x}{2}}+C.\end{aligned}}$

Fractions

In general, this technique may be used to evaluate any fractions involving trigonometric functions. For example, consider the integral

$\int {\frac {1+\cos ^{2}x}{\cos x+\cos 3x}}\,dx.$

Using Euler's identity, this integral becomes

${\frac {1}{2}}\int {\frac {6+e^{2ix}+e^{-2ix}}{e^{ix}+e^{-ix}+e^{3ix}+e^{-3ix}}}\,dx.$

If we now make the substitution u = e^ix, the result is the integral of a rational function:

${\frac {i}{2}}\int {\frac {1+6u^{2}+u^{4}}{1+u^{2}+u^{4}+u^{6}}}\,du.$

Any rational function is integrable (using, for example, partial fractions), and therefore any fraction involving trigonometric functions may be integrated as well.

External links

• Evaluation of difficult integrals using complex numbers and Euler's formula(Broken)