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시컨트 함수의 적분

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시컨트 함수(붉은색)과 그 부정적분(푸른색)의 그래프

미적분학에서, 시컨트 함수의 적분은 다양한 방법으로 계산할 수 있고 부정적분을 표현하는 방법도 여러 가지이지만, 이들 모두 삼각함수 항등식에 의해 동치로 나타낼 수 있다.

This formula is useful for evaluating various trigonometric integrals. In particular, it can be used to evaluate the integral of the secant cubed, which, though seemingly special, comes up rather frequently in applications.[1]

The definite integral of the secant function starting from is the inverse Gudermannian function, For numerical applications, all of the above expressions result in loss of significance for some arguments. An alternative expression in terms of the inverse hyperbolic sine 틀:수학/style.css 문서에 내용이 없습니다.arsinh is numerically well behaved for real arguments :[2]

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. 시컨트 함수의 에서 시작하는 정적분은 구데르만 함수의 역함수 이다.

Proof that the different antiderivatives are equivalent[편집]

삼각함수 형태[편집]

시컨트 함수의 적분의 세 가지 범용적인 표현

는 다음과 같은 이유로 동치이다:

증명: 세 식에 각각 탄젠트 반각 치환 를 적용하여 에 대해 동일하게 표현되는 것을 보이면 된다. 이 치환을 적용하면 , 이다.

첫째,

둘째,

셋째, 탄젠트의 덧셈정리 를 적용하면,

따라서 세 식은 모두 같다.

The conventional solution for the Mercator projection ordinate may be written without the absolute value signs since the latitude lies between and ,

쌍곡선 형태[편집]

라 한다. 그러면,

History[편집]

  1. 넘겨주기 틀:광범위

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] He wanted the solution for the purposes of cartography – specifically for constructing an accurate Mercator projection.[3] In the 1640s, Henry Bond, a teacher of navigation, surveying, and other mathematical topics, compared Wright's numerically computed table of values of the integral of the secant with a table of logarithms of the tangent function, and consequently conjectured that[3]

This conjecture became widely known, and in 1665, Isaac Newton was aware of it.[5]

계산[편집]

표준적인 치환(그레고리의 접근)[편집]

A standard method of evaluating the secant integral presented in various references involves multiplying the numerator and denominator by 틀:수학/style.css 문서에 내용이 없습니다.sec θ + tan θ and then using the substitution 틀:수학/style.css 문서에 내용이 없습니다.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]

Starting with

adding them gives

The derivative of the sum is thus equal to the sum multiplied by 틀:수학/style.css 문서에 내용이 없습니다.sec θ. This enables multiplying 틀:수학/style.css 문서에 내용이 없습니다.sec θ by 틀:수학/style.css 문서에 내용이 없습니다.sec θ + tan θ in the numerator and denominator and performing the following substitutions:

The integral is evaluated as follows:

as claimed. This was the formula discovered by James Gregory.[1]

부분분수와 치환(배로의 접근)[편집]

Although Gregory proved the conjecture in 1668 in his Exercitationes Geometricae,[7] the proof was presented in a form that renders it nearly impossible for modern readers to comprehend; Isaac Barrow, in his Lectiones Geometricae of 1670,[8] gave the first "intelligible" proof, though even that was "couched in the geometric idiom of the day."[3] Barrow's proof of the result was the earliest use of partial fractions in integration.[3] Adapted to modern notation, Barrow's proof began as follows:

Substituting 틀:수학/style.css 문서에 내용이 없습니다.u = sin θ, 틀:수학/style.css 문서에 내용이 없습니다.du = cos θ , reduces the integral to

Therefore,

as expected. Taking the absolute value is not necessary because and are always non-negative for real values of

탄젠트 반각 치환[편집]

표준적인 방식[편집]

Under the tangent half-angle substitution [9]

Therefore the integral of the secant function is

as before.

비표준적인 방식[편집]

The integral can also be derived by using a somewhat non-standard version of the tangent half-angle substitution, which is simpler in the case of this particular integral, published in 2013,[10] is as follows:

Substituting:

2번 연속으로 치환[편집]

The integral can also be solved by manipulating the integrand and substituting twice. Using the definition 틀:수학/style.css 문서에 내용이 없습니다.sec θ =

  1. 넘겨주기 틀:수직분수 and the identity 틀:수학/style.css 문서에 내용이 없습니다.cos2θ + sin2θ = 1, the integral can be rewritten as

Substituting 틀:수학/style.css 문서에 내용이 없습니다.u = sin θ, 틀:수학/style.css 문서에 내용이 없습니다.du = cos θ reduces the integral to

The reduced integral can be evaluated by substituting 틀:수학/style.css 문서에 내용이 없습니다.u = tanh t, 틀:수학/style.css 문서에 내용이 없습니다.du = sech2t dt, and then using the identity 틀:수학/style.css 문서에 내용이 없습니다.1 − tanh2t = sech2t.

The integral is now reduced to a simple integral, and back-substituting gives

which is one of the hyperbolic forms of the integral.

A similar strategy can be used to integrate the cosecant, hyperbolic secant, and hyperbolic cosecant functions.

기타 쌍곡선 형태[편집]

It is also possible to find the other two hyperbolic forms directly, by again multiplying and dividing by a convenient term:

where stands for because Substituting 틀:수학/style.css 문서에 내용이 없습니다.u = tan θ, 틀:수학/style.css 문서에 내용이 없습니다.du = sec2θ , reduces to a standard integral:

where 틀:수학/style.css 문서에 내용이 없습니다.sgn is the sign function.

Likewise:

Substituting 틀:수학/style.css 문서에 내용이 없습니다.u =

  1. 넘겨주기 틀:절댓값, 틀:수학/style.css 문서에 내용이 없습니다.du =
  2. 넘겨주기 틀:절댓값 tan θ , reduces to a standard integral:

복소지수함수 형태[편집]

Under the substitution

So the integral can be solved as:

Because the constant of integration can be anything, the additional constant term can be absorbed into it. Finally, if theta is real-valued, we can indicate this with absolute value brackets in order to get the equation into its most familiar form:

구데르만과 람베르트 함수[편집]

The Gudermannian function relates the area of a circular sector to the area of a hyperbolic sector, via a common stereographic projection. If twice the area of the blue hyperbolic sector is 틀:수학/style.css 문서에 내용이 없습니다.ψ, then twice the area of the red circular sector is 틀:수학/style.css 문서에 내용이 없습니다.ϕ = gd ψ. Twice the area of the purple triangle is the stereographic projection 틀:수학/style.css 문서에 내용이 없습니다.s = tan  #넘겨주기 틀:수직분수ϕ = tanh  #넘겨주기 틀:수직분수ψ. The blue point has coordinates 틀:수학/style.css 문서에 내용이 없습니다.(cosh ψ, sinh ψ). The red point has coordinates 틀:수학/style.css 문서에 내용이 없습니다.(cos ϕ, sin ϕ). The purple point has coordinates 틀:수학/style.css 문서에 내용이 없습니다.(0, s).

The integral of the hyperbolic secant function defines the Gudermannian function:

The integral of the secant function defines the Lambertian function, which is the inverse of the Gudermannian function:

These functions are encountered in the theory of map projections: the Mercator projection of a point on the sphere with longitude 틀:수학/style.css 문서에 내용이 없습니다.λ and latitude 틀:수학/style.css 문서에 내용이 없습니다.ϕ may be written[11] as:

더 보기[편집]

package.lua 80번째 줄에서 Lua 오류: module 'Module:Portal/그림/기타' not found.

노트[편집]

  1. 1.0 1.1 1.2 스크립트 오류: "citation/CS1" 모듈이 없습니다.
  2. For example this form is used in 스크립트 오류: "Citation/CS1" 모듈이 없습니다.
  3. 3.0 3.1 3.2 3.3 3.4 V. Frederick Rickey and Philip M. Tuchinsky, An Application of Geography to Mathematics: History of the Integral of the Secant in Mathematics Magazine, volume 53, number 3, May 1980, pages 162–166.
  4. Edward Wright, Certaine Errors in Navigation, Arising either of the ordinaire erroneous making or vsing of the sea Chart, Compasse, Crosse staffe, and Tables of declination of the Sunne, and fixed Starres detected and corrected, Valentine Simms, London, 1599.
  5. H. W. Turnbull, editor, The Correspondence of Isaac Newton, Cambridge University Press, 1959–1960, volume 1, pages 13–16 and volume 2, pages 99–100.

    D. T. Whiteside, editor, The Mathematical Papers of Isaac Newton, Cambridge University Press, 1967, volume 1, pages 466–467 and 473–475.

  6. 스크립트 오류: "citation/CS1" 모듈이 없습니다.

    스크립트 오류: "citation/CS1" 모듈이 없습니다.

  7. 스크립트 오류: "citation/CS1" 모듈이 없습니다.
  8. 스크립트 오류: "citation/CS1" 모듈이 없습니다. In English, 스크립트 오류: "citation/CS1" 모듈이 없습니다.
  9. 스크립트 오류: "citation/CS1" 모듈이 없습니다.
  10. 스크립트 오류: "Citation/CS1" 모듈이 없습니다.
  11. 스크립트 오류: "citation/CS1" 모듈이 없습니다.

각주[편집]

  • 스크립트 오류: "citation/CS1" 모듈이 없습니다.
  • 스크립트 오류: "Citation/CS1" 모듈이 없습니다.



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