weierstrass substitution proof

The content of PM is described in a section by section synopsis, stated in modernized logical notation and described following the introductory notes from each of the three . The Weierstrass substitution formulas are most useful for integrating rational functions of sine and cosine (http://planetmath.org/IntegrationOfRationalFunctionOfSineAndCosine). What is a word for the arcane equivalent of a monastery? \text{tan}x&=\frac{2u}{1-u^2} \\ d This is really the Weierstrass substitution since $t=\tan(x/2)$. weierstrass substitution proof. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. |Contents| In trigonometry, tangent half-angle formulas relate the tangent of half of an angle to trigonometric functions of the entire angle. ) "1.4.6. The best answers are voted up and rise to the top, Not the answer you're looking for? This is the discriminant. Our aim in the present paper is twofold. Define: b 2 = a 1 2 + 4 a 2. b 4 = 2 a 4 + a 1 a 3. b 6 = a 3 2 + 4 a 6. b 8 = a 1 2 a 6 + 4 a 2 a 6 a 1 a 3 a 4 + a 2 a 3 2 a 4 2. Now for a given > 0 there exist > 0 by the definition of uniform continuity of functions. &=\text{ln}|u|-\frac{u^2}{2} + C \\ Now he could get the area of the blue region because sector $CPQ^{\prime}$ of the circle centered at $C$, at $-ae$ on the $x$-axis and radius $a$ has area $$\frac12a^2E$$ where $E$ is the eccentric anomaly and triangle $COQ^{\prime}$ has area $$\frac12ae\cdot\frac{a\sqrt{1-e^2}\sin\nu}{1+e\cos\nu}=\frac12a^2e\sin E$$ so the area of blue sector $OPQ^{\prime}$ is $$\frac12a^2(E-e\sin E)$$ Why do small African island nations perform better than African continental nations, considering democracy and human development? = Here we shall see the proof by using Bernstein Polynomial. The Weierstrass Approximation theorem is named after German mathematician Karl Theodor Wilhelm Weierstrass. \end{align} \theta = 2 \arctan\left(t\right) \implies in his 1768 integral calculus textbook,[3] and Adrien-Marie Legendre described the general method in 1817. Stack Exchange network consists of 181 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. sin The complete edition of Bolzano's works (Bernard-Bolzano-Gesamtausgabe) was founded by Jan Berg and Eduard Winter together with the publisher Gnther Holzboog, and it started in 1969.Since then 99 volumes have already appeared, and about 37 more are forthcoming. sines and cosines can be expressed as rational functions of Let \(K\) denote the field we are working in. Mayer & Mller. Finally, it must be clear that, since \(\text{tan}x\) is undefined for \(\frac{\pi}{2}+k\pi\), \(k\) any integer, the substitution is only meaningful when restricted to intervals that do not contain those values, e.g., for \(-\pi\lt x\lt\pi\). Hyperbolic Tangent Half-Angle Substitution, Creative Commons Attribution/Share-Alike License, https://mathworld.wolfram.com/WeierstrassSubstitution.html, https://proofwiki.org/w/index.php?title=Weierstrass_Substitution&oldid=614929, $\mathsf{Pr} \infty \mathsf{fWiki}$ $\LaTeX$ commands, Creative Commons Attribution-ShareAlike License, Weisstein, Eric W. "Weierstrass Substitution." x Karl Weierstrass, in full Karl Theodor Wilhelm Weierstrass, (born Oct. 31, 1815, Ostenfelde, Bavaria [Germany]died Feb. 19, 1897, Berlin), German mathematician, one of the founders of the modern theory of functions. transformed into a Weierstrass equation: We only consider cubic equations of this form. &= \frac{\sec^2 \frac{x}{2}}{(a + b) + (a - b) \tan^2 \frac{x}{2}}, Linear Algebra - Linear transformation question. &=\text{ln}|\text{tan}(x/2)|-\frac{\text{tan}^2(x/2)}{2} + C. [2] Leonhard Euler used it to evaluate the integral Is it known that BQP is not contained within NP? He is best known for the Casorati Weierstrass theorem in complex analysis. Then the integral is written as. t = The Weierstrass Approximation theorem To perform the integral given above, Kepler blew up the picture by a factor of $1/\sqrt{1-e^2}$ in the $y$-direction to turn the ellipse into a circle. are well known as Weierstrass's inequality [1] or Weierstrass's Bernoulli's inequality [3]. and the integral reads {\textstyle x=\pi } By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Thus, dx=21+t2dt. gives, Taking the quotient of the formulae for sine and cosine yields. as follows: Using the double-angle formulas, introducing denominators equal to one thanks to the Pythagorean theorem, and then dividing numerators and denominators by 1. u File history. tan The method is known as the Weierstrass substitution. \(\text{cos}\theta=\frac{BC}{AB}=\frac{1-u^2}{1+u^2}\). It is also assumed that the reader is familiar with trigonometric and logarithmic identities. {\displaystyle t} d \). The equation for the drawn line is y = (1 + x)t. The equation for the intersection of the line and circle is then a quadratic equation involving t. The two solutions to this equation are (1, 0) and (cos , sin ). Yet the fascination of Dirichlet's Principle itself persisted: time and again attempts at a rigorous proof were made. 1 t . We generally don't use the formula written this w.ay oT do a substitution, follow this procedure: Step 1 : Choose a substitution u = g(x). doi:10.1007/1-4020-2204-2_16. tan The technique of Weierstrass Substitution is also known as tangent half-angle substitution . The simplest proof I found is on chapter 3, "Why Does The Miracle Substitution Work?" "8. If you do use this by t the power goes to 2n. b Weierstrass's theorem has a far-reaching generalizationStone's theorem. Proof by contradiction - key takeaways. {\textstyle x} What can a lawyer do if the client wants him to be acquitted of everything despite serious evidence? Since, if 0 f Bn(x, f) and if g f Bn(x, f). where $a$ and $e$ are the semimajor axis and eccentricity of the ellipse. = The plots above show for (red), 3 (green), and 4 (blue). d x x t $\qquad$ $\endgroup$ - Michael Hardy for both limits of integration. 2 The point. Vol. The Weierstrass substitution, named after German mathematician Karl Weierstrass (18151897), is used for converting rational expressions of trigonometric functions into algebraic rational functions, which may be easier to integrate. \implies follows is sometimes called the Weierstrass substitution. These two answers are the same because If an integrand is a function of only \(\tan x,\) the substitution \(t = \tan x\) converts this integral into integral of a rational function. t Viewed 270 times 2 $\begingroup$ After browsing some topics here, through one post, I discovered the "miraculous" Weierstrass substitutions. = $$\begin{align}\int\frac{dx}{a+b\cos x}&=\frac1a\int\frac{d\nu}{1+e\cos\nu}=\frac12\frac1{\sqrt{1-e^2}}\int dE\\ Date/Time Thumbnail Dimensions User (d) Use what you have proven to evaluate R e 1 lnxdx. \end{align} \implies & d\theta = (2)'\!\cdot\arctan\left(t\right) + 2\!\cdot\!\big(\arctan\left(t\right)\big)' 5. . Note sur l'intgration de la fonction, https://archive.org/details/coursdanalysedel01hermuoft/page/320/, https://archive.org/details/anelementarytre00johngoog/page/n66, https://archive.org/details/traitdanalyse03picagoog/page/77, https://archive.org/details/courseinmathemat01gouruoft/page/236, https://archive.org/details/advancedcalculus00wils/page/21/, https://archive.org/details/treatiseonintegr01edwauoft/page/188, https://archive.org/details/ost-math-courant-differentialintegralcalculusvoli/page/n250, https://archive.org/details/elementsofcalcul00pete/page/201/, https://archive.org/details/calculus0000apos/page/264/, https://archive.org/details/calculuswithanal02edswok/page/482, https://archive.org/details/calculusofsingle00lars/page/520, https://books.google.com/books?id=rn4paEb8izYC&pg=PA435, https://books.google.com/books?id=R-1ZEAAAQBAJ&pg=PA409, "The evaluation of trigonometric integrals avoiding spurious discontinuities", "A Note on the History of Trigonometric Functions", https://en.wikipedia.org/w/index.php?title=Tangent_half-angle_substitution&oldid=1137371172, This page was last edited on 4 February 2023, at 07:50. A standard way to calculate \(\int{\frac{dx}{1+\text{sin}x}}\) is via a substitution \(u=\text{tan}(x/2)\). 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). Kluwer. cos It is sometimes misattributed as the Weierstrass substitution. & \frac{\theta}{2} = \arctan\left(t\right) \implies x As x varies, the point (cosx,sinx) winds repeatedly around the unit circle centered at(0,0). 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). WEIERSTRASS APPROXIMATION THEOREM TL welll kroorn Neiendsaas . = Definition of Bernstein Polynomial: If f is a real valued function defined on [0, 1], then for n N, the nth Bernstein Polynomial of f is defined as . . . (This is the one-point compactification of the line.) u Redoing the align environment with a specific formatting. Published by at 29, 2022. Disconnect between goals and daily tasksIs it me, or the industry. doi:10.1145/174603.174409. Thus, when Weierstrass found a flaw in Dirichlet's Principle and, in 1869, published his objection, it . t cot . H. Anton, though, warns the student that the substitution can lead to cumbersome partial fractions decompositions and consequently should be used only in the absence of finding a simpler method. / No clculo integral, a substituio tangente do arco metade ou substituio de Weierstrass uma substituio usada para encontrar antiderivadas e, portanto, integrais definidas, de funes racionais de funes trigonomtricas.Nenhuma generalidade perdida ao considerar que essas so funes racionais do seno e do cosseno. assume the statement is false). We show how to obtain the difference function of the Weierstrass zeta function very directly, by choosing an appropriate order of summation in the series defining this function. 2 Another way to get to the same point as C. Dubussy got to is the following: Thus, the tangent half-angle formulae give conversions between the stereographic coordinate t on the unit circle and the standard angular coordinate . In the year 1849, C. Hermite first used the notation 123 for the basic Weierstrass doubly periodic function with only one double pole. The Weierstrass approximation theorem. As t goes from 1 to0, the point follows the part of the circle in the fourth quadrant from (0,1) to(1,0). Also, using the angle addition and subtraction formulae for both the sine and cosine one obtains: Pairwise addition of the above four formulae yields: Setting must be taken into account. It's not difficult to derive them using trigonometric identities. {\textstyle t=\tan {\tfrac {x}{2}},} Weierstrass Approximation theorem in real analysis presents the notion of approximating continuous functions by polynomial functions. $$\int\frac{dx}{a+b\cos x}=\frac1a\int\frac{dx}{1+\frac ba\cos x}=\frac1a\int\frac{d\nu}{1+\left|\frac ba\right|\cos\nu}$$ Merlet, Jean-Pierre (2004). Finally, fifty years after Riemann, D. Hilbert . csc Try to generalize Additional Problem 2. cos csc By eliminating phi between the directly above and the initial definition of 2 Is there a single-word adjective for "having exceptionally strong moral principles"? 1 Introducing a new variable This equation can be further simplified through another affine transformation. Here we shall see the proof by using Bernstein Polynomial. This is helpful with Pythagorean triples; each interior angle has a rational sine because of the SAS area formula for a triangle and has a rational cosine because of the Law of Cosines. So if doing an integral with a factor of $\frac1{1+e\cos\nu}$ via the eccentric anomaly was good enough for Kepler, surely it's good enough for us. So to get $\nu(t)$, you need to solve the integral The best answers are voted up and rise to the top, Not the answer you're looking for? When $a,b=1$ we can just multiply the numerator and denominator by $1-\cos x$ and that solves the problem nicely. for \(\mathrm{char} K \ne 2\), we have that if \((x,y)\) is a point, then \((x, -y)\) is In other words, if f is a continuous real-valued function on [a, b] and if any > 0 is given, then there exist a polynomial P on [a, b] such that |f(x) P(x)| < , for every x in [a, b]. The proof of this theorem can be found in most elementary texts on real . Alternatively, first evaluate the indefinite integral, then apply the boundary values. it is, in fact, equivalent to the completeness axiom of the real numbers. How to solve the integral $\int\limits_0^a {\frac{{\sqrt {{a^2} - {x^2}} }}{{b - x}}} \mathop{\mathrm{d}x}\\$? identities (see Appendix C and the text) can be used to simplify such rational expressions once we make a preliminary substitution. It uses the substitution of u= tan x 2 : (1) The full method are substitutions for the values of dx, sinx, cosx, tanx, cscx, secx, and cotx. {\textstyle \int dx/(a+b\cos x)} 382-383), this is undoubtably the world's sneakiest substitution. The above descriptions of the tangent half-angle formulae (projection the unit circle and standard hyperbola onto the y-axis) give a geometric interpretation of this function. two values that \(Y\) may take. Integration of Some Other Classes of Functions 13", "Intgration des fonctions transcendentes", "19. 2 tan This method of integration is also called the tangent half-angle substitution as it implies the following half-angle identities: where \(t = \tan \frac{x}{2}\) or \(x = 2\arctan t.\). or a singular point (a point where there is no tangent because both partial cot {\displaystyle 1+\tan ^{2}\alpha =1{\big /}\cos ^{2}\alpha } 4. cos Polynomial functions are simple functions that even computers can easily process, hence the Weierstrass Approximation theorem has great practical as well as theoretical utility. By similarity of triangles. Use the universal trigonometric substitution: \[dx = d\left( {2\arctan t} \right) = \frac{{2dt}}{{1 + {t^2}}}.\], \[{\cos ^2}x = \frac{1}{{1 + {{\tan }^2}x}} = \frac{1}{{1 + {t^2}}},\;\;\;{\sin ^2}x = \frac{{{{\tan }^2}x}}{{1 + {{\tan }^2}x}} = \frac{{{t^2}}}{{1 + {t^2}}}.\], \[t = \tan \frac{x}{2},\;\; \Rightarrow x = 2\arctan t,\;\;\; dx = \frac{{2dt}}{{1 + {t^2}}}.\], \[\int {\frac{{dx}}{{1 + \sin x}}} = \int {\frac{{\frac{{2dt}}{{1 + {t^2}}}}}{{1 + \frac{{2t}}{{1 + {t^2}}}}}} = \int {\frac{{2dt}}{{1 + {t^2} + 2t}}} = \int {\frac{{2dt}}{{{{\left( {t + 1} \right)}^2}}}} = - \frac{2}{{t + 1}} + C = - \frac{2}{{\tan \frac{x}{2} + 1}} + C.\], \[x = \arctan t,\;\; \sin x = \frac{{2t}}{{1 + {t^2}}},\;\; dx = \frac{{2dt}}{{1 + {t^2}}},\], \[I = \int {\frac{{dx}}{{3 - 2\sin x}}} = \int {\frac{{\frac{{2dt}}{{1 + {t^2}}}}}{{3 - 2 \cdot \frac{{2t}}{{1 + {t^2}}}}}} = \int {\frac{{2dt}}{{3 + 3{t^2} - 4t}}} = \int {\frac{{2dt}}{{3\left( {{t^2} - \frac{4}{3}t + 1} \right)}}} = \frac{2}{3}\int {\frac{{dt}}{{{t^2} - \frac{4}{3}t + 1}}} .\], \[{t^2} - \frac{4}{3}t + 1 = {t^2} - \frac{4}{3}t + {\left( {\frac{2}{3}} \right)^2} - {\left( {\frac{2}{3}} \right)^2} + 1 = {\left( {t - \frac{2}{3}} \right)^2} - \frac{4}{9} + 1 = {\left( {t - \frac{2}{3}} \right)^2} + \frac{5}{9} = {\left( {t - \frac{2}{3}} \right)^2} + {\left( {\frac{{\sqrt 5 }}{3}} \right)^2}.\], \[I = \frac{2}{3}\int {\frac{{dt}}{{{{\left( {t - \frac{2}{3}} \right)}^2} + {{\left( {\frac{{\sqrt 5 }}{3}} \right)}^2}}}} = \frac{2}{3}\int {\frac{{du}}{{{u^2} + {{\left( {\frac{{\sqrt 5 }}{3}} \right)}^2}}}} = \frac{2}{3} \cdot \frac{1}{{\frac{{\sqrt 5 }}{3}}}\arctan \frac{u}{{\frac{{\sqrt 5 }}{3}}} + C = \frac{2}{{\sqrt 5 }}\arctan \frac{{3\left( {t - \frac{2}{3}} \right)}}{{\sqrt 5 }} + C = \frac{2}{{\sqrt 5 }}\arctan \frac{{3t - 2}}{{\sqrt 5 }} + C = \frac{2}{{\sqrt 5 }}\arctan \left( {\frac{{3\tan \frac{x}{2} - 2}}{{\sqrt 5 }}} \right) + C.\], \[t = \tan \frac{x}{4},\;\; \Rightarrow d\left( {\frac{x}{2}} \right) = \frac{{2dt}}{{1 + {t^2}}},\;\; \Rightarrow \cos \frac{x}{2} = \frac{{1 - {t^2}}}{{1 + {t^2}}}.\], \[\int {\frac{{dx}}{{1 + \cos \frac{x}{2}}}} = \int {\frac{{d\left( {\frac{x}{2}} \right)}}{{1 + \cos \frac{x}{2}}}} = 2\int {\frac{{\frac{{2dt}}{{1 + {t^2}}}}}{{1 + \frac{{1 - {t^2}}}{{1 + {t^2}}}}}} = 4\int {\frac{{dt}}{{1 + \cancel{t^2} + 1 - \cancel{t^2}}}} = 2\int {dt} = 2t + C = 2\tan \frac{x}{4} + C.\], \[t = \tan x,\;\; \Rightarrow x = \arctan t,\;\; \Rightarrow dx = \frac{{dt}}{{1 + {t^2}}},\;\; \Rightarrow \cos 2x = \frac{{1 - {t^2}}}{{1 + {t^2}}},\], \[\int {\frac{{dx}}{{1 + \cos 2x}}} = \int {\frac{{\frac{{dt}}{{1 + {t^2}}}}}{{1 + \frac{{1 - {t^2}}}{{1 + {t^2}}}}}} = \int {\frac{{dt}}{{1 + \cancel{t^2} + 1 - \cancel{t^2}}}} = \int {\frac{{dt}}{2}} = \frac{t}{2} + C = \frac{1}{2}\tan x + C.\], \[t = \tan \frac{x}{4},\;\; \Rightarrow x = 4\arctan t,\;\; dx = \frac{{4dt}}{{1 + {t^2}}},\;\; \cos \frac{x}{2} = \frac{{1 - {t^2}}}{{1 + {t^2}}}.\], \[\int {\frac{{dx}}{{4 + 5\cos \frac{x}{2}}}} = \int {\frac{{\frac{{4dt}}{{1 + {t^2}}}}}{{4 + 5 \cdot \frac{{1 - {t^2}}}{{1 + {t^2}}}}}} = \int {\frac{{4dt}}{{4\left( {1 + {t^2}} \right) + 5\left( {1 - {t^2}} \right)}}} = 4\int {\frac{{dt}}{{4 + 4{t^2} + 5 - 5{t^2}}}} = 4\int {\frac{{dt}}{{{3^2} - {t^2}}}} = 4 \cdot \frac{1}{{2 \cdot 3}}\ln \left| {\frac{{3 + t}}{{3 - t}}} \right| + C = \frac{2}{3}\ln \left| {\frac{{3 + \tan \frac{x}{4}}}{{3 - \tan \frac{x}{4}}}} \right| + C.\], \[\int {\frac{{dx}}{{\sin x + \cos x}}} = \int {\frac{{\frac{{2dt}}{{1 + {t^2}}}}}{{\frac{{2t}}{{1 + {t^2}}} + \frac{{1 - {t^2}}}{{1 + {t^2}}}}}} = \int {\frac{{2dt}}{{2t + 1 - {t^2}}}} = 2\int {\frac{{dt}}{{1 - \left( {{t^2} - 2t} \right)}}} = 2\int {\frac{{dt}}{{1 - \left( {{t^2} - 2t + 1 - 1} \right)}}} = 2\int {\frac{{dt}}{{2 - {{\left( {t - 1} \right)}^2}}}} = 2\int {\frac{{d\left( {t - 1} \right)}}{{{{\left( {\sqrt 2 } \right)}^2} - {{\left( {t - 1} \right)}^2}}}} = 2 \cdot \frac{1}{{2\sqrt 2 }}\ln \left| {\frac{{\sqrt 2 + \left( {t - 1} \right)}}{{\sqrt 2 - \left( {t - 1} \right)}}} \right| + C = \frac{1}{{\sqrt 2 }}\ln \left| {\frac{{\sqrt 2 - 1 + \tan \frac{x}{2}}}{{\sqrt 2 + 1 - \tan \frac{x}{2}}}} \right| + C.\], \[t = \tan \frac{x}{2},\;\; \Rightarrow x = 2\arctan t,\;\; dx = \frac{{2dt}}{{1 + {t^2}}},\;\; \sin x = \frac{{2t}}{{1 + {t^2}}},\;\; \cos x = \frac{{1 - {t^2}}}{{1 + {t^2}}}.\], \[\int {\frac{{dx}}{{\sin x + \cos x + 1}}} = \int {\frac{{\frac{{2dt}}{{1 + {t^2}}}}}{{\frac{{2t}}{{1 + {t^2}}} + \frac{{1 - {t^2}}}{{1 + {t^2}}} + 1}}} = \int {\frac{{\frac{{2dt}}{{1 + {t^2}}}}}{{\frac{{2t + 1 - {t^2} + 1 + {t^2}}}{{1 + {t^2}}}}}} = \int {\frac{{2dt}}{{2t + 2}}} = \int {\frac{{dt}}{{t + 1}}} = \ln \left| {t + 1} \right| + C = \ln \left| {\tan \frac{x}{2} + 1} \right| + C.\], \[I = \int {\frac{{dx}}{{\sec x + 1}}} = \int {\frac{{dx}}{{\frac{1}{{\cos x}} + 1}}} = \int {\frac{{\cos xdx}}{{1 + \cos x}}} .\], \[I = \int {\frac{{\cos xdx}}{{1 + \cos x}}} = \int {\frac{{\frac{{1 - {t^2}}}{{1 + {t^2}}} \cdot \frac{{2dt}}{{1 + {t^2}}}}}{{1 + \frac{{1 - {t^2}}}{{1 + {t^2}}}}}} = 2\int {\frac{{\frac{{1 - {t^2}}}{{{{\left( {1 + {t^2}} \right)}^2}}}dt}}{{\frac{{1 + {t^2} + 1 - {t^2}}}{{1 + {t^2}}}}}} = \int {\frac{{1 - {t^2}}}{{1 + {t^2}}}dt} = - \int {\frac{{1 + {t^2} - 2}}{{1 + {t^2}}}dt} = - \int {1dt} + 2\int {\frac{{dt}}{{1 + {t^2}}}} = - t + 2\arctan t + C = - \tan \frac{x}{2} + 2\arctan \left( {\tan \frac{x}{2}} \right) + C = x - \tan \frac{x}{2} + C.\], Trigonometric and Hyperbolic Substitutions. = It yields: The Weierstrass substitution is an application of Integration by Substitution . ) = \), \( Ask Question Asked 7 years, 9 months ago. $$ This is the \(j\)-invariant. Geometrical and cinematic examples. The Weierstrass substitution is an application of Integration by Substitution. by setting Finding $\int \frac{dx}{a+b \cos x}$ without Weierstrass substitution. \text{cos}x&=\frac{1-u^2}{1+u^2} \\ csc 2 [5] It is known in Russia as the universal trigonometric substitution,[6] and also known by variant names such as half-tangent substitution or half-angle substitution. If \(a_1 = a_3 = 0\) (which is always the case x File:Weierstrass substitution.svg. Die Weierstra-Substitution ist eine Methode aus dem mathematischen Teilgebiet der Analysis. Multivariable Calculus Review. , ( The Weierstrass substitution, named after German mathematician Karl Weierstrass (18151897), is used for converting rational expressions of trigonometric functions into algebraic rational functions, which may be easier to integrate.. https://mathworld.wolfram.com/WeierstrassSubstitution.html. With the objective of identifying intrinsic forms of mathematical production in complex analysis (CA), this study presents an analysis of the mathematical activity of five original works that . How to handle a hobby that makes income in US. An irreducibe cubic with a flex can be affinely x and substituting yields: Dividing the sum of sines by the sum of cosines one arrives at: Applying the formulae derived above to the rhombus figure on the right, it is readily shown that. The integral on the left is $-\cot x$ and the one on the right is an easy $u$-sub with $u=\sin x$. \begin{aligned} &=\int{\frac{2du}{1+2u+u^2}} \\ . \end{aligned} , (1/2) The tangent half-angle substitution relates an angle to the slope of a line. These inequalities are two o f the most important inequalities in the supject of pro duct polynomials. Solution. The formulation throughout was based on theta functions, and included much more information than this summary suggests. into one of the following forms: (Im not sure if this is true for all characteristics.). x Sie ist eine Variante der Integration durch Substitution, die auf bestimmte Integranden mit trigonometrischen Funktionen angewendet werden kann. . This method of integration is also called the tangent half-angle substitution as it implies the following half-angle identities: