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These identities are useful whenever expressions involving trigonometric functions need to be simplified. An important application is the integration of non-trigonometric functions: a common technique involves first using the substitution rule with a trigonometric function, and then simplifying the resulting integral with a trigonometric identity.
For the special antiderivatives involving trigonometric functions, see Trigonometric integral. [ 1 ] Generally, if the function sin x {\displaystyle \sin x} is any trigonometric function, and cos x {\displaystyle \cos x} is its derivative,
Since sinc is an even entire function (holomorphic over the entire complex plane), Si is entire, odd, and the integral in its definition can be taken along any path connecting the endpoints. By definition, Si(x) is the antiderivative of sin x / x whose value is zero at x = 0, and si(x) is the antiderivative whose value is zero at x = ∞.
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 and and then integrated.
That is, ω acts like an even function. This is the same as the symmetry of the cosine, which is an even function, so the mnemonic tells us to use the substitution = (rule 1). Under this substitution, the integral becomes . The integrand involving transcendental functions has been reduced to one involving a rational function (a constant).
If units of degrees are intended, the degree sign must be explicitly shown (sin x°, cos x°, etc.). Using this standard notation, the argument x for the trigonometric functions satisfies the relationship x = (180 x / π )°, so that, for example, sin π = sin 180° when we take x = π .
For a definite integral, one must figure out how the bounds of integration change. For example, as x {\displaystyle x} goes from 0 {\displaystyle 0} to a / 2 , {\displaystyle a/2,} then sin θ {\displaystyle \sin \theta } goes from 0 {\displaystyle 0} to 1 / 2 , {\displaystyle 1/2,} so θ {\displaystyle \theta } goes from 0 {\displaystyle 0 ...
This visualization also explains why integration by parts may help find the integral of an inverse function f −1 (x) when the integral of the function f(x) is known. Indeed, the functions x(y) and y(x) are inverses, and the integral ∫ x dy may be calculated as above from knowing the integral ∫ y dx.