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Integration using Euler's formula
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<h2 id="eulers-formula">Euler's formula</h2> <p>Euler's formula states that<a class="footnote-ref" id="fnref:2" href="#fn:2"><sup>2</sup></a> </p> e i x = cos x + i sin x . {\displaystyle e^{ix}=\cos x+i\,\sin x.} <p>Substituting − x {\displaystyle -x} for x {\displaystyle x} gives the equation </p> e − i x = cos x − i sin x {\displaystyle e^{-ix}=\cos x-i\,\sin x} <p>because cosine is an even function and sine is odd. These two equations can be solved for the sine and cosine to give </p> 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}}.} <h2 id="examples">Examples</h2> <h3>First example</h3> <p>Consider the integral </p> ∫ cos 2 x d x . {\displaystyle \int \cos ^{2}x\,dx.} <p>The standard approach to this integral is to use a <a href="/facts/Half-angle_formula/Xw9Jx8Vz">half-angle formula</a> to simplify the integrand. We can use Euler's identity instead: </p> ∫ 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}}} <p>At this point, it would be possible to change back to real numbers using the formula <i>e</i>2<i>ix</i> + <i>e</i>−2<i>ix</i> = 2 cos 2<i>x</i>. Alternatively, we can integrate the complex exponentials and not change back to trigonometric functions until the end: </p> 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}}} <h3>Second example</h3> <p>Consider the integral </p> ∫ sin 2 x cos 4 x d x . {\displaystyle \int \sin ^{2}x\cos 4x\,dx.} <p>This integral would be extremely tedious to solve using trigonometric identities, but using Euler's identity makes it relatively painless: </p> ∫ 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}}} <p>At this point we can either integrate directly, or we can first change the integrand to 2 cos 6<i>x</i> − 4 cos 4<i>x</i> + 2 cos 2<i>x</i> and continue from there. Either method gives </p> ∫ 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.} <h2 id="using-real-parts">Using real parts</h2> <p>In addition to Euler's identity, it can be helpful to make judicious use of the <a href="/facts/Real_part/2w7mqI7e">real parts</a> of complex expressions. For example, consider the integral </p> ∫ e x cos x d x . {\displaystyle \int e^{x}\cos x\,dx.} <p>Since cos <i>x</i> is the real part of <i>e</i><i>ix</i>, we know that </p> ∫ 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.} <p>The integral on the right is easy to evaluate: </p> ∫ 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.} <p>Thus: </p> ∫ 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}}} <h2 id="fractions">Fractions</h2> <p>In general, this technique may be used to evaluate any fractions involving trigonometric functions. For example, consider the integral </p> ∫ 1 + cos 2 x cos x + cos 3 x d x . {\displaystyle \int {\frac {1+\cos ^{2}x}{\cos x+\cos 3x}}\,dx.} <p>Using Euler's identity, this integral becomes </p> 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.} <p>If we now make the <a href="/facts/Integration_by_substitution/NvXfTTaf">substitution</a> u = e i x {\displaystyle u=e^{ix}} , the result is the integral of a <a href="/facts/Rational_function/2nJGu0Ko">rational function</a>: </p> − 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.} <p>One may proceed using <a href="/facts/Partial_fraction_decomposition/KLRmM3Hw">partial fraction decomposition</a>. </p> <h2 id="see-also">See also</h2> <ul> <li>Mathematics portal</li></ul> <ul><li><a href="/facts/Trigonometric_substitution/U6LBLAYU">Trigonometric substitution</a></li> <li><a href="/facts/Weierstrass_substitution/3up24fUw">Weierstrass substitution</a></li> <li><a href="/facts/Euler_substitution/oN3tJSDb">Euler substitution</a></li></ul> <h2 id="references">References</h2> <ol> <li id="fn:1"><p>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) <a href="https://arms.thebrpi.org/vol-7-no-2-december-2019-abstract-1-arms" target="_blank">https://arms.thebrpi.org/vol-7-no-2-december-2019-abstract-1-arms</a> <a href="#fnref:1" class="footnote-back-ref">↩</a></p></li> <li id="fn:2"><p>Weisstein, Eric W. "Euler Formula". mathworld.wolfram.com. Retrieved 2021-03-17. <a href="https://mathworld.wolfram.com/EulerFormula.html" target="_blank">https://mathworld.wolfram.com/EulerFormula.html</a> <a href="#fnref:2" class="footnote-back-ref">↩</a></p></li> </ol>