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For antiderivatives involving both exponential and trigonometric functions, see List of integrals of exponential functions. For a complete list of antiderivative functions, see Lists of integrals. For the special antiderivatives involving trigonometric functions, see Trigonometric integral. [1]
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
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 = ∞.
The integral of the secant function was one of the "outstanding open problems of the mid-seventeenth century", solved in 1668 by James Gregory. [3] He applied his result to a problem concerning nautical tables. [1] In 1599, Edward Wright evaluated the integral by numerical methods – what today we would call Riemann sums. [4]
The figure at the right shows a sector of a circle with radius 1. The sector is θ/(2 π) of the whole circle, so its area is θ/2. We assume here that θ < π /2. = = = = The area of triangle OAD is AB/2, or sin(θ)/2.
In this right triangle, denoting the measure of angle BAC as A: sin A = a / c ; cos A = b / c ; tan A = a / b . Plot of the six trigonometric functions, the unit circle, and a line for the angle θ = 0.7 radians. The points labeled 1, Sec(θ), Csc(θ) represent the length of the line segment from the origin to that point.
An even larger, multivolume table is the Integrals and Series by Prudnikov, Brychkov, and Marichev (with volumes 1–3 listing integrals and series of elementary and special functions, volume 4–5 are tables of Laplace transforms).
As t goes from 0 to 1, the point follows the part of the circle in the first quadrant from (1, 0) to (0, 1). Finally, as t goes from 1 to +∞, the point follows the part of the circle in the second quadrant from (0, 1) to (−1, 0). Here is another geometric point of view. Draw the unit circle, and let P be the point (−1, 0).