Euler's formula states that ^[2]

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

Substituting ${\displaystyle -x}$  for ${\displaystyle x}$  gives the equation

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

because cosine is an even function and sine is odd. These two equations can be solved for the sine and cosine to give

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

First example edit

Consider the integral

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

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

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

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:

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

Second example edit

Consider the integral

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

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

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

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

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

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

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

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

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

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:

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

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.

• Trigonometric substitution
• Weierstrass substitution
• Euler substitution