Integration using Euler's formula

Examples

First example

Consider the integral $${\displaystyle \int {\cos }^{2}xdx.}$$ The standard approach to this integral is to use a half-angle formula to simplify the integrand. We can use Euler's identity instead: $${\displaystyle \begin{array}{cc}\int {\cos }^{2}xdx& =\int {\left(\frac{e^{ix}+e^{-ix}}{2}\right)}^{2}dx\\ & =\frac{1}{4}\int \left(e^{2ix}+2+e^{-2ix}\right)dx\end{array}}$$ 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: $${\displaystyle \begin{array}{cc}\frac{1}{4}\int \left(e^{2ix}+2+e^{-2ix}\right)dx& =\frac{1}{4}\left(\frac{e^{2ix}}{2i}+2x-\frac{e^{-2ix}}{2i}\right)+C\\ & =\frac{1}{4}\left(2x+\sin 2x\right)+C.\end{array}}$$

Second example

Consider the integral $${\displaystyle \int {\sin }^{2}x\cos 4xdx.}$$ This integral would be extremely tedious to solve using trigonometric identities, but using Euler's identity makes it relatively painless: $${\displaystyle \begin{array}{cc}\int {\sin }^{2}x\cos 4xdx& =\int {\left(\frac{e^{ix}-e^{-ix}}{2i}\right)}^2\left(\frac{e^{4ix}+e^{-4ix}}{2}\right)dx\\ & =-\frac{1}{8}\int \left(e^{2ix}-2+e^{-2ix}\right)\left(e^{4ix}+e^{-4ix}\right)dx\\ & =-\frac{1}{8}\int \left(e^{6ix}-2e^{4ix}+e^{2ix}+e^{-2ix}-2e^{-4ix}+e^{-6ix}\right)dx.\end{array}}$$ At this point we can either integrate directly, or we can first change the integrand to 2 cos 6x − 4 cos 4x + 2 cos 2x and continue from there. Either method gives $${\displaystyle \int {\sin }^{2}x\cos 4xdx=-\frac{1}{24}\sin 6x+\frac{1}{8}\sin 4x-\frac{1}{8}\sin 2x+C.}$$

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 $${\displaystyle \int e^x\cos xdx.}$$ Since cos x is the real part of e^ix, we know that $${\displaystyle \int e^x\cos xdx=\mathrm{Re}\int e^x{e}^{ix}dx.}$$ The integral on the right is easy to evaluate: $${\displaystyle \int e^x{e}^{ix}dx=\int e^{(1+i)x}dx=\frac{e^{(1+i)x}}{1+i}+C.}$$ Thus: $${\displaystyle \begin{array}{cc}\int e^x\cos xdx& =\mathrm{Re}\left(\frac{e^{(1+i)x}}{1+i}\right)+C\\ & =e^x\mathrm{Re}\left(\frac{e^{ix}}{1+i}\right)+C\\ & =e^x\mathrm{Re}\left(\frac{e^{ix}(1-i)}{2}\right)+C\\ & =e^x\frac{\cos x+\sin x}{2}+C.\end{array}}$$

Fractions

In general, this technique may be used to evaluate any fractions involving trigonometric functions. For example, consider the integral $${\displaystyle \int \frac{1+{\cos }^{2}x}{\cos x+\cos 3x}dx.}$$ Using Euler's identity, this integral becomes $${\displaystyle \frac{1}{2}\int \frac{6+e^{2ix}+e^{-2ix}}{e^{ix}+e^{-ix}+e^{3ix}+e^{-3ix}}dx.}$$ If we now make the substitution ${\displaystyle u=e^{ix}}$, the result is the integral of a rational function: $${\displaystyle -\frac{i}{2}\int \frac{1+6u^2+u^4}{1+u^2+u^4+u^6}du.}$$ One may proceed using partial fraction decomposition.