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The integral of secant cubed is a frequent and challenging [1] indefinite integral of elementary calculus:
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]
Then | | = (()) +, where sgn(x) is the sign function, which takes the values −1, 0, 1 when x is respectively negative, zero or positive. This can be proved by computing the derivative of the right-hand side of the formula, taking into account that the condition on g is here for insuring the continuity of the integral.
(Note that the value of the expression is independent of the value of n, which is why it does not appear in the integral.) ∫ x x ⋅ ⋅ x ⏟ m d x = ∑ n = 0 m ( − 1 ) n ( n + 1 ) n − 1 n !
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.
Just as the definite integral of a positive function of one variable represents the area of the region between the graph of the function and the x-axis, the double integral of a positive function of two variables represents the volume of the region between the surface defined by the function (on the three-dimensional Cartesian plane where z = f(x, y)) and the plane which contains its domain. [1]
3.1 Integrals of hyperbolic tangent, cotangent, secant, cosecant functions 3.2 Integrals involving hyperbolic sine and cosine functions 3.3 Integrals involving hyperbolic and trigonometric functions
Integrands of the form (d + e x) m (a + b x + c x 2) p when b 2 − 4 a c = 0 [ edit ] The resulting integrands are of the same form as the original integrand, so these reduction formulas can be repeatedly applied to drive the exponents m and p toward 0.