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| | Hi! <br>My name is Edgardo and I'm a 25 years old girl from Italy.<br>xunjie コンピュータを自動的に開いて、 |
| The '''[[integral]] of the [[Trigonometric_functions#Reciprocal_functions|secant function]]''' of [[trigonometry]] was the subject of one of the "outstanding open problems of the mid-seventeenth century", solved in 1668 by [[James Gregory (mathematician)|James Gregory]].<ref name="rickey-tuchinsky">V. Frederick Rickey and Philip M. Tuchinsky, "An Application of Geography to Mathematics: History of the Integral of the Secant", ''[[Mathematics Magazine]]'', volume 53, number 3, May 1980, pages 162–166.</ref> In 1599, [[Edward Wright (mathematician)|Edward Wright]] evaluated the integral by [[numerical method]]s – what today we would call [[Riemann sum]]s.<ref>[[Edward Wright (mathematician)|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.</ref> He wanted the solution for the purposes of [[cartography]] – specifically for constructing an accurate [[Mercator projection]].<ref name="rickey-tuchinsky"/> 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<ref name="rickey-tuchinsky"/> that
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| : <math> \int_0^\theta \sec\zeta\,d\zeta = \ln\left|\tan\left(\frac{\theta}{2} + \frac{\pi}{4}\right)\right|. </math>
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| That conjecture became widely known, and in 1665, [[Isaac Newton]] was aware of it.<ref>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.</ref><ref>[[D. T. Whiteside]], editor, ''The Mathematical Papers of Isaac Newton'', Cambridge University Press, 1967, volume 1, pages 466–467 and 473–475.</ref>
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| The problem was solved by [[Isaac Barrow]]. His proof of the result was the earliest use of [[partial fraction]]s in integration.<ref name="rickey-tuchinsky"/> Adapted to modern notation, Barrow's proof began as follows:
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| : <math>
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| \int \sec \theta \, d\theta = \int \frac{d\theta}{\cos\theta} = \int \frac{\cos\theta \, d\theta}{\cos^2\theta} = \int \frac{\cos\theta \, d\theta}{1 - \sin^2\theta} = \int \frac{du}{1 - u^2}
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| </math>
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| This reduces it to the problem of antidifferentiating a [[rational function]] by using partial fractions. The proof goes on from there:
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| : <math>
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| \begin{align}
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| \int \frac{du}{1 - u^2} & = \int\frac{du}{(1-u)(1+u)} = \dfrac12\int \left(\frac{1}{1+u} + \frac{1}{1-u}\right)\,du \\[10pt]
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| & = \frac12 \ln \left|1 + u\right| - \frac12 \ln \left|1 - u\right| + C = \frac12 \ln\left|\frac{1+u}{1-u}\right| + C
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| \end{align}
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| </math>
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| Finally, we convert it back to a function of ''θ'':
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| : <math>
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| = \left\{\begin{array}{l}
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| \dfrac12 \ln \left|\dfrac{1+\sin\theta}{1-\sin\theta}\right| + C \\[15pt]
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| \ln\left|\sec\theta + \tan\theta\right| + C \\[15pt]
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| \ln\left| \tan\left(\dfrac{\theta}{2} + \dfrac{\pi}{4}\right) \right| + C
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| \end{array}\right\}\text{ (equivalent forms)}
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| </math>
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| The third form may be obtained directly by means of the following substitutions.
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| : <math>
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| \begin{align}
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| \sec\theta=\frac{1}{\sin\left(\theta + \dfrac{\pi}{2}\right)}
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| =\frac{1}{2\sin\left(\dfrac{\theta}{2} + \dfrac{\pi}{4}\right)
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| \cos\left(\dfrac{\theta}{2} + \dfrac{\pi}{4}\right)}
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| =\frac{\sec^2\left(\dfrac{\theta}{2} + \dfrac{\pi}{4}\right)}
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| {2\tan\left(\dfrac{\theta}{2} + \dfrac{\pi}{4}\right)}.
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| \end{align}
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| </math>
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| The conventional solution for the [[Mercator projection]] ordinate may be written without the modulus signs since the latitude (φ) lies between −π/2 and π/2:
| |
| :<math>
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| y= \ln \tan\!\left(\dfrac{\phi}{2} + \dfrac{\pi}{4}\right).
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| </math>
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| The problem can also be done by using the [[tangent half-angle substitution]], but the details become somewhat more complicated than in the argument above.
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| ==Hyperbolic forms==
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| Let
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| :<math>
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| \begin{align}
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| \psi &=\ln(\sec\theta+\tan\theta),\\
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| {\rm e}^\psi &=\sec\theta+\tan\theta,\\
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| \sinh\psi &=\frac12({\rm e}^\psi-{\rm e}^{-\psi})=\tan\theta,\\
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| \cosh\psi &=\sqrt{1+\sinh^2\psi}=\sec\theta,\\
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| \tanh\psi &=\sin\theta.
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| \end{align}
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| </math>
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| Therefore
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| :<math>
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| \begin{align}
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| \int \sec \theta \, d\theta&
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| =\tanh^{-1}\! \left(\sin\theta\right)
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| =\sinh^{-1}\! \left(\tan\theta\right)
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| =\cosh^{-1}\! \left(\sec\theta\right).
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| \end{align}
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| </math>
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| ==Gudermannian and lambertian==
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| :<math>
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| \begin{align}
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| \int \sec \theta \, d\theta& = \mbox{gd}^{-1}(\theta)=\mbox{lam}(\theta).
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| \end{align}
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| </math>
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| gd is the [[Gudermannian function]].
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| The lambertian form (lam) is encountered in the theory of map projections.<ref name=lee_exact>Lee, L.P. (1976). ''Conformal Projections Based on Elliptic Functions''. Supplement No. 1 to Canadian Cartographer, Vol 13. (Designated as Monograph 16)</ref>
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| == Notes and references ==
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| {{reflist}}
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| == See also ==
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| * [[Integral of secant cubed]]
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| * [[Gudermannian function]]
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| == External links ==
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| * [http://www.jstor.org/stable/2690106 Rickey and Tuchinsky's paper on the history of this integral]
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| [[Category:Integral calculus]]
| |
Hi!
My name is Edgardo and I'm a 25 years old girl from Italy.
xunjie コンピュータを自動的に開いて、
117.46パーセントで1044000ドルに475000ドルから増加し成長しながら、
赤ちゃんの両親がインストールされ、 [http://www.tobler-verlag.ch/media/galerie/jp/shop/nb/ �˥�`�Х�� �����ȥ�å�] 1969年に起源を何年も何年後にこのブランドはなってきている、
これらは普通の人々は場所を見ることができないです。
数百の直接支援万ものがあるタレントコンテストに参加して美しい赤ちゃんのジュエリーは、 [http://www.zgg.ch/data/video/shop/gaga.html �����ߥ�� �rӋ ��ǥ��`�� �ԥ�] ゴールデン真珠のブローチマルティーヌWester3。
小さな家族を得意とすることを推奨し、
細い超とを連れて行かれたの冷艳妖艶匂いを、[http://athenavineyards.com/media/duvetica.php �ޥå����ީ`�� �����`�������] 私は私の服を扱うだろうかと考えると、
日本市場のリーダーでの販売の長い期間、
特に活発なスポーツアリーナになびく、
編集（上）ボッテガ·ヴェネタ2012年春と夏のファッションのリリース：ルフトハンザspinulosaのchawol時間：2011年9月25日 [http://athenavineyards.com/media/duvetica.php �ǥ�٥
� ��ǥ��`��]
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