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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,
Si(x) (blue) and Ci(x) (green) shown on the same plot. Integral sine in the complex plane, plotted with a variant of domain coloring. Integral cosine in the complex plane. Note the branch cut along the negative real axis. In mathematics, trigonometric integrals are a family of nonelementary integrals involving trigonometric functions.
A formula for computing the trigonometric identities for the one-third angle exists, but it requires finding the zeroes of the cubic equation 4x 3 − 3x + d = 0, where is the value of the cosine function at the one-third angle and d is the known value of the cosine function at
If the function f does not have any continuous antiderivative which takes the value zero at the zeros of f (this is the case for the sine and the cosine functions), then sgn(f(x)) ∫ f(x) dx is an antiderivative of f on every interval on which f is not zero, but may be discontinuous at the points where f(x) = 0.
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 e i x {\displaystyle e^{ix}} and e − i x {\displaystyle e^{-ix}} and then integrated.
For a definite integral, the bounds change once the substitution is performed and are determined using the equation = , with values in the range < <. Alternatively, apply the boundary terms directly to the formula for the antiderivative.
A line integral (sometimes called a path integral) is an integral where the function to be integrated is evaluated along a curve. [42] Various different line integrals are in use. In the case of a closed curve it is also called a contour integral. The function to be integrated may be a scalar field or a vector field.
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.