In integral calculus, Euler's formula for complex numbers may be used to evaluate integrals involving trigonometric functions. Using Euler's formula, any trigonometric function may be written in terms of complex exponential functions, namely ${\displaystyle e^{ix}}$ and ${\displaystyle e^{-ix}}$ 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.^[1]

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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

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

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