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At this point we can either integrate directly, or we can first change the integrand to 2 cos 6x − 4 cos 4x + 2 cos 2x and continue from there. Either method gives Either method gives ∫ 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 ...
The following is a list of integrals (antiderivative functions) of trigonometric functions.For antiderivatives involving both exponential and trigonometric functions, see List of integrals of exponential functions.
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 = ∞.
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.
For a complete list of integral functions, see list of integrals. In all formulas the constant a is assumed to be nonzero, and C denotes the constant of integration . Integrals involving only hyperbolic sine functions
The integral of the secant function was historically one of the first integrals of its type ever evaluated, before most of the development of integral calculus. It is important because it is the vertical coordinate of the Mercator projection , used for marine navigation with constant compass bearing .
The slope field of () = +, showing three of the infinitely many solutions that can be produced by varying the arbitrary constant c.. In calculus, an antiderivative, inverse derivative, primitive function, primitive integral or indefinite integral [Note 1] of a continuous function f is a differentiable function F whose derivative is equal to the original function f.
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.