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Inintegral calculus,Euler's formula forcomplex numbers may be used to evaluateintegrals involvingtrigonometric 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 usingtrigonometric identities orintegration by parts, and is sufficiently powerful to integrate anyrational expression involving trigonometric functions.^[1]

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

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

Consider the integral

${\displaystyle \int {\cos }^{2}xdx.}$

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

At this point, it would be possible to change back to real numbers using the formulae^2ix +e^−2ix = 2 cos 2x. Alternatively, we can integrate the complex exponentials and not change back to trigonometric functions until the end:

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:

At this point we can either integrate directly, or we can first change the integrand to2 cos 6x − 4 cos 4x + 2 cos 2x and continue from there.Either method gives

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

In addition to Euler's identity, it can be helpful to make judicious use of thereal parts of complex expressions. For example, consider the integral

${\displaystyle \int e^x\cos xdx.}$

Sincecosx is the real part ofe^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:

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 thesubstitution${\displaystyle u=e^{ix}}$, the result is the integral of arational function:

${\displaystyle -\frac{i}{2}\int \frac{1+6u^2+u^4}{1+u^2+u^4+u^6}du.}$

One may proceed usingpartial fraction decomposition.

• Trigonometric substitution
• Weierstrass substitution
• Euler substitution

• Integral calculus
• Theorems in mathematical analysis
• Theorems in calculus

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