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Just as the integration by parts above reduced the integral of secant cubed to the integral of secant to the first power, so a similar process reduces the integral of higher odd powers of secant to lower ones. This is the secant reduction formula, which follows the syntax:
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. This gives the following formulas (where a ≠ 0), which are valid over any interval where f is continuous (over larger intervals, the constant C must be replaced ...
A standard method of evaluating the secant integral presented in various references involves multiplying the numerator and denominator by sec θ + tan θ and then using the substitution u = sec θ + tan θ. This substitution can be obtained from the derivatives of secant and tangent added together, which have secant as a common factor. [6]
In mathematics, the definite integral ()is the area of the region in the xy-plane bounded by the graph of f, the x-axis, and the lines x = a and x = b, such that area above the x-axis adds to the total, and that below the x-axis subtracts from the total.
Indefinite integrals are antiderivative functions. A constant (the constant of integration) may be added to the right hand side of any of these formulas, but has been suppressed here in the interest of brevity.
The following is a list of integrals (anti-derivative functions) of hyperbolic functions. 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.
These reduction formulas can be used for integrands having integer and/or fractional exponents. Special cases of these reductions formulas can be used for integrands of the form ( a + b x n + c x 2 n ) p {\displaystyle \left(a+b\,x^{n}+c\,x^{2n}\right)^{p}} when b 2 − 4 a c = 0 {\displaystyle b^{2}-4\,a\,c=0} by setting m to 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.