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Euler's formula

Euler's formula states that²

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

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

e − i x = cos ⁡ x − i sin ⁡ x {\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

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

Examples

First example

Consider the integral

∫ cos 2 ⁡ x d x . {\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:

∫ cos 2 ⁡ x d x = ∫ ( e i x + e − i x 2 ) 2 d x = 1 4 ∫ ( e 2 i x + 2 + e − 2 i x ) d x {\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 e2ix + e−2ix = 2 cos 2x. Alternatively, we can integrate the complex exponentials and not change back to trigonometric functions until the end:

1 4 ∫ ( e 2 i x + 2 + e − 2 i x ) d x = 1 4 ( e 2 i x 2 i + 2 x − e − 2 i x 2 i ) + C = 1 4 ( 2 x + sin ⁡ 2 x ) + C . {\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

Consider the integral

∫ sin 2 ⁡ x cos ⁡ 4 x d x . {\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:

∫ sin 2 ⁡ x cos ⁡ 4 x d x = ∫ ( e i x − e − i x 2 i ) 2 ( e 4 i x + e − 4 i x 2 ) d x = − 1 8 ∫ ( e 2 i x − 2 + e − 2 i x ) ( e 4 i x + e − 4 i x ) d x = − 1 8 ∫ ( e 6 i x − 2 e 4 i x + e 2 i x + e − 2 i x − 2 e − 4 i x + e − 6 i x ) d x . {\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

∫ sin 2 ⁡ x cos ⁡ 4 x d x = − 1 24 sin ⁡ 6 x + 1 8 sin ⁡ 4 x − 1 8 sin ⁡ 2 x + C . {\displaystyle \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

∫ e x cos ⁡ x d x . {\displaystyle \int e^{x}\cos x\,dx.}

Since cos x is the real part of eix, we know that

∫ e x cos ⁡ x d x = Re ⁡ ∫ e x e i x d x . {\displaystyle \int e^{x}\cos x\,dx=\operatorname {Re} \int e^{x}e^{ix}\,dx.}

The integral on the right is easy to evaluate:

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

Thus:

∫ e x cos ⁡ x d x = Re ⁡ ( e ( 1 + i ) x 1 + i ) + C = e x Re ⁡ ( e i x 1 + i ) + C = e x Re ⁡ ( e i x ( 1 − i ) 2 ) + C = e x cos ⁡ x + sin ⁡ x 2 + C . {\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}}}

Fractions

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

∫ 1 + cos 2 ⁡ x cos ⁡ x + cos ⁡ 3 x d x . {\displaystyle \int {\frac {1+\cos ^{2}x}{\cos x+\cos 3x}}\,dx.}

Using Euler's identity, this integral becomes

1 2 ∫ 6 + e 2 i x + e − 2 i x e i x + e − i x + e 3 i x + e − 3 i x d x . {\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 u = e i x {\displaystyle u=e^{ix}} , the result is the integral of a rational function:

− i 2 ∫ 1 + 6 u 2 + u 4 1 + u 2 + u 4 + u 6 d u . {\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.

See also

• Mathematics portal

• Trigonometric substitution
• Weierstrass substitution
• Euler substitution

References

1. Kilburn, Korey (2019). "Applying Euler's Formula to Integrate". American Review of Mathematics and Statistics. 7. American Research Institute for Policy Development: 1–2. doi:10.15640/arms.v7n2a1 (inactive 1 November 2024). eISSN 2374-2356. hdl:2158/1183208. ISSN 2374-2348.{{cite journal}}: CS1 maint: DOI inactive as of November 2024 (link) https://arms.thebrpi.org/vol-7-no-2-december-2019-abstract-1-arms ↩

2. Weisstein, Eric W. "Euler Formula". mathworld.wolfram.com. Retrieved 2021-03-17. https://mathworld.wolfram.com/EulerFormula.html ↩