The Weierstrass representation is particularly useful for constructing immersed minimal surfaces. Learn more about Stack Overflow the company, and our products. , rearranging, and taking the square roots yields. Theorems on differentiation, continuity of differentiable functions. u In integral calculus, the tangent half-angle substitution - known in Russia as the universal trigonometric substitution, sometimes misattributed as the Weierstrass substitution, and also known by variant names such as half-tangent substitution or half-angle substitution - is a change of variables used for evaluating integrals, which converts a rational function of trigonometric functions . 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. Vol. File history. Why are physically impossible and logically impossible concepts considered separate in terms of probability? into one of the form. x What is the correct way to screw wall and ceiling drywalls? \text{tan}x&=\frac{2u}{1-u^2} \\ $\qquad$. The Weierstrass substitution is an application of Integration by Substitution . With or without the absolute value bars these formulas do not apply when both the numerator and denominator on the right-hand side are zero. sin t x ( An irreducibe cubic with a flex can be affinely As x varies, the point (cos x . If \(a_1 = a_3 = 0\) (which is always the case "8. That is often appropriate when dealing with rational functions and with trigonometric functions. x ISBN978-1-4020-2203-6. Is it correct to use "the" before "materials used in making buildings are"? Styling contours by colour and by line thickness in QGIS. The Weierstrass Substitution The Weierstrass substitution enables any rational function of the regular six trigonometric functions to be integrated using the methods of partial fractions. = That is often appropriate when dealing with rational functions and with trigonometric functions. Why do academics stay as adjuncts for years rather than move around? I saw somewhere on Math Stack that there was a way of finding integrals in the form $$\int \frac{dx}{a+b \cos x}$$ without using Weierstrass substitution, which is the usual technique. Karl Theodor Wilhelm Weierstrass ; 1815-1897 . According to the theorem, every continuous function defined on a closed interval [a, b] can approximately be represented by a polynomial function. By the Stone Weierstrass Theorem we know that the polynomials on [0,1] [ 0, 1] are dense in C ([0,1],R) C ( [ 0, 1], R). Is there a single-word adjective for "having exceptionally strong moral principles"? Evaluate the integral \[\int {\frac{{dx}}{{1 + \sin x}}}.\], Evaluate the integral \[\int {\frac{{dx}}{{3 - 2\sin x}}}.\], Calculate the integral \[\int {\frac{{dx}}{{1 + \cos \frac{x}{2}}}}.\], Evaluate the integral \[\int {\frac{{dx}}{{1 + \cos 2x}}}.\], Compute the integral \[\int {\frac{{dx}}{{4 + 5\cos \frac{x}{2}}}}.\], Find the integral \[\int {\frac{{dx}}{{\sin x + \cos x}}}.\], Find the integral \[\int {\frac{{dx}}{{\sin x + \cos x + 1}}}.\], Evaluate \[\int {\frac{{dx}}{{\sec x + 1}}}.\]. This allows us to write the latter as rational functions of t (solutions are given below). My code is GPL licensed, can I issue a license to have my code be distributed in a specific MIT licensed project? 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It only takes a minute to sign up. $$\int\frac{dx}{a+b\cos x}=\frac1a\int\frac{dx}{1+\frac ba\cos x}=\frac1a\int\frac{d\nu}{1+\left|\frac ba\right|\cos\nu}$$ A little lowercase underlined 'u' character appears on your Site design / logo 2023 Stack Exchange Inc; user contributions licensed under CC BY-SA. According to Spivak (2006, pp. Other sources refer to them merely as the half-angle formulas or half-angle formulae. The Bolzano Weierstrass theorem is named after mathematicians Bernard Bolzano and Karl Weierstrass. Thus, dx=21+t2dt. By Weierstrass Approximation Theorem, there exists a sequence of polynomials pn on C[0, 1], that is, continuous functions on [0, 1], which converges uniformly to f. Since the given integral is convergent, we have. Especially, when it comes to polynomial interpolations in numerical analysis. Integrate $\int \frac{\sin{2x}}{\sin{x}+\cos^2{x}}dx$, Find the indefinite integral $\int \frac{25}{(3\cos(x)+4\sin(x))^2} dx$. It only takes a minute to sign up. The Bernstein Polynomial is used to approximate f on [0, 1]. d . p 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 . $$\int\frac{d\nu}{(1+e\cos\nu)^2}$$ t = When $a,b=1$ we can just multiply the numerator and denominator by $1-\cos x$ and that solves the problem nicely. where gd() is the Gudermannian function. This is very useful when one has some process which produces a " random " sequence such as what we had in the idea of the alleged proof in Theorem 7.3. (c) Finally, use part b and the substitution y = f(x) to obtain the formula for R b a f(x)dx. $$. cos pp. Integration by substitution to find the arc length of an ellipse in polar form. 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). t The plots above show for (red), 3 (green), and 4 (blue). + The Weierstrass substitution is the trigonometric substitution which transforms an integral of the form. 2 You can still apply for courses starting in 2023 via the UCAS website. Fact: The discriminant is zero if and only if the curve is singular. p at Browse other questions tagged, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site. . Try to generalize Additional Problem 2. The Weierstrass substitution in REDUCE. {\textstyle \cos ^{2}{\tfrac {x}{2}},} The secant integral may be evaluated in a similar manner. ( |Contents| . 2.4: The Bolazno-Weierstrass Theorem - Mathematics LibreTexts Connect and share knowledge within a single location that is structured and easy to search. Using the above formulas along with the double angle formulas, we obtain, sinx=2sin(x2)cos(x2)=2t1+t211+t2=2t1+t2. The Bolzano-Weierstrass Theorem is at the foundation of many results in analysis. Alternatively, first evaluate the indefinite integral, then apply the boundary values. Why do academics stay as adjuncts for years rather than move around? t arbor park school district 145 salary schedule; Tags . \implies &\bbox[4pt, border:1.25pt solid #000000]{d\theta = \frac{2\,dt}{1 + t^{2}}} A place where magic is studied and practiced? = If so, how close was it? Preparation theorem. 1 In Weierstrass form, we see that for any given value of \(X\), there are at most The tangent half-angle substitution parametrizes the unit circle centered at (0, 0). Here is another geometric point of view. Tangent line to a function graph. Here we shall see the proof by using Bernstein Polynomial. Did any DOS compatibility layers exist for any UNIX-like systems before DOS started to become outmoded? 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. For an even and $2\pi$ periodic function, why does $\int_{0}^{2\pi}f(x)dx = 2\int_{0}^{\pi}f(x)dx $. 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. The proof of this theorem can be found in most elementary texts on real . . Bibliography. 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. In the case = 0, we get the well-known perturbation theory for the sine-Gordon equation. "The evaluation of trigonometric integrals avoiding spurious discontinuities". 20 (1): 124135. {\textstyle u=\csc x-\cot x,} 2.1.5Theorem (Weierstrass Preparation Theorem)Let U A V A Fn Fbe a neighbourhood of (x;0) and suppose that the holomorphic or real analytic function A . Why is there a voltage on my HDMI and coaxial cables? The Weierstrass substitution parametrizes the unit circle centered at (0, 0). {\textstyle t=\tan {\tfrac {x}{2}}} http://www.westga.edu/~faucette/research/Miracle.pdf, We've added a "Necessary cookies only" option to the cookie consent popup, Integrating trig substitution triangle equivalence, Elementary proof of Bhaskara I's approximation: $\sin\theta=\frac{4\theta(180-\theta)}{40500-\theta(180-\theta)}$, Weierstrass substitution on an algebraic expression. Denominators with degree exactly 2 27 . These identities can be useful in calculus for converting rational functions in sine and cosine to functions of t in order to find their antiderivatives. \theta = 2 \arctan\left(t\right) \implies https://mathworld.wolfram.com/WeierstrassSubstitution.html. Some sources call these results the tangent-of-half-angle formulae. (1/2) The tangent half-angle substitution relates an angle to the slope of a line. , 4. Substitute methods had to be invented to . Kluwer. weierstrass substitution proof. cot b |Front page| Linear Algebra - Linear transformation question. cot . \end{align} t It applies to trigonometric integrals that include a mixture of constants and trigonometric function. If we identify the parameter t in both cases we arrive at a relationship between the circular functions and the hyperbolic ones. An irreducibe cubic with a flex can be affinely transformed into a Weierstrass equation: Y 2 + a 1 X Y + a 3 Y = X 3 + a 2 X 2 + a 4 X + a 6. Following this path, we are able to obtain a system of differential equations that shows the amplitude and phase modulation of the approximate solution. , differentiation rules imply. In various applications of trigonometry, it is useful to rewrite the trigonometric functions (such as sine and cosine) in terms of rational functions of a new variable &=\int{\frac{2du}{(1+u)^2}} \\ \text{sin}x&=\frac{2u}{1+u^2} \\ {\textstyle \int dx/(a+b\cos x)} Note that $$\frac{1}{a+b\cos(2y)}=\frac{1}{a+b(2\cos^2(y)-1)}=\frac{\sec^2(y)}{2b+(a-b)\sec^2(y)}=\frac{\sec^2(y)}{(a+b)+(a-b)\tan^2(y)}.$$ Hence $$\int \frac{dx}{a+b\cos(x)}=\int \frac{\sec^2(y)}{(a+b)+(a-b)\tan^2(y)} \, dy.$$ Now conclude with the substitution $t=\tan(y).$, Kepler found the substitution when he was trying to solve the equation {\textstyle \csc x-\cot x=\tan {\tfrac {x}{2}}\colon }. \), \( Categories . . 382-383), this is undoubtably the world's sneakiest substitution. But here is a proof without words due to Sidney Kung: \(\text{sin}\theta=\frac{AC}{AB}=\frac{2u}{1+u^2}\) and The technique of Weierstrass Substitution is also known as tangent half-angle substitution. {\displaystyle t} Benannt ist die Methode nach dem Mathematiker Karl Weierstra, der sie entwickelte. Instead of a closed bounded set Rp, we consider a compact space X and an algebra C ( X) of continuous real-valued functions on X. {\displaystyle t} Later authors, citing Stewart, have sometimes referred to this as the Weierstrass substitution, for instance: Jeffrey, David J.; Rich, Albert D. (1994). \begin{aligned} This paper studies a perturbative approach for the double sine-Gordon equation. as follows: Using the double-angle formulas, introducing denominators equal to one thanks to the Pythagorean theorem, and then dividing numerators and denominators by 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." and Die Weierstra-Substitution ist eine Methode aus dem mathematischen Teilgebiet der Analysis. What is the correct way to screw wall and ceiling drywalls? The Bolzano-Weierstrass Property and Compactness. Integrate $\int \frac{4}{5+3\cos(2x)}\,d x$. Thus, when Weierstrass found a flaw in Dirichlet's Principle and, in 1869, published his objection, it . Typically, it is rather difficult to prove that the resulting immersion is an embedding (i.e., is 1-1), although there are some interesting cases where this can be done. \( of this paper: http://www.westga.edu/~faucette/research/Miracle.pdf. It's not difficult to derive them using trigonometric identities. Define: \(b_8 = a_1^2 a_6 + 4a_2 a_6 - a_1 a_3 a_4 + a_2 a_3^2 - a_4^2\). {\textstyle t=\tan {\tfrac {x}{2}}} $\int\frac{a-b\cos x}{(a^2-b^2)+b^2(\sin^2 x)}dx$. Proof of Weierstrass Approximation Theorem . Weierstrass, Karl (1915) [1875]. This is really the Weierstrass substitution since $t=\tan(x/2)$. Do roots of these polynomials approach the negative of the Euler-Mascheroni constant? By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. and a rational function of \end{align} In the first line, one cannot simply substitute How can Kepler know calculus before Newton/Leibniz were born ? 382-383), this is undoubtably the world's sneakiest substitution. WEIERSTRASS APPROXIMATION THEOREM TL welll kroorn Neiendsaas . The sigma and zeta Weierstrass functions were introduced in the works of F . Published by at 29, 2022. Mayer & Mller. The function was published by Weierstrass but, according to lectures and writings by Kronecker and Weierstrass, Riemann seems to have claimed already in 1861 that . 2 The Gudermannian function gives a direct relationship between the circular functions and the hyperbolic ones that does not involve complex numbers. Did this satellite streak past the Hubble Space Telescope so close that it was out of focus? 2 Weierstrass Substitution 24 4. can be expressed as the product of = Projecting this onto y-axis from the center (1, 0) gives the following: Finding in terms of t leads to following relationship between the inverse hyperbolic tangent x x 1. To calculate an integral of the form \(\int {R\left( {\sin x} \right)\cos x\,dx} ,\) where \(R\) is a rational function, use the substitution \(t = \sin x.\), Similarly, to calculate an integral of the form \(\int {R\left( {\cos x} \right)\sin x\,dx} ,\) where \(R\) is a rational function, use the substitution \(t = \cos x.\). \). & \frac{\theta}{2} = \arctan\left(t\right) \implies , cosx=cos2(x2)-sin2(x2)=(11+t2)2-(t1+t2)2=11+t2-t21+t2=1-t21+t2. + Required fields are marked *, \(\begin{array}{l}\sum_{k=0}^{n}f\left ( \frac{k}{n} \right )\begin{pmatrix}n \\k\end{pmatrix}x_{k}(1-x)_{n-k}\end{array} \), \(\begin{array}{l}\sum_{k=0}^{n}(f-f(\zeta))\left ( \frac{k}{n} \right )\binom{n}{k} x^{k}(1-x)^{n-k}\end{array} \), \(\begin{array}{l}\sum_{k=0}^{n}\binom{n}{k}x^{k}(1-x)^{n-k} = (x+(1-x))^{n}=1\end{array} \), \(\begin{array}{l}\left|B_{n}(x, f)-f(\zeta) \right|=\left|B_{n}(x,f-f(\zeta)) \right|\end{array} \), \(\begin{array}{l}\leq B_{n}\left ( x,2M\left ( \frac{x- \zeta}{\delta } \right )^{2}+ \frac{\epsilon}{2} \right ) \end{array} \), \(\begin{array}{l}= \frac{2M}{\delta ^{2}} B_{n}(x,(x- \zeta )^{2})+ \frac{\epsilon}{2}\end{array} \), \(\begin{array}{l}B_{n}(x, (x- \zeta)^{2})= x^{2}+ \frac{1}{n}(x x^{2})-2 \zeta x + \zeta ^{2}\end{array} \), \(\begin{array}{l}\left| (B_{n}(x,f)-f(\zeta))\right|\leq \frac{\epsilon}{2}+\frac{2M}{\delta ^{2}}(x- \zeta)^{2}+\frac{2M}{\delta^{2}}\frac{1}{n}(x- x ^{2})\end{array} \), \(\begin{array}{l}\left| (B_{n}(x,f)-f(\zeta))\right|\leq \frac{\epsilon}{2}+\frac{2M}{\delta ^{2}}\frac{1}{n}(\zeta- \zeta ^{2})\end{array} \), \(\begin{array}{l}\left| (B_{n}(x,f)-f(\zeta))\right|\leq \frac{\epsilon}{2}+\frac{M}{2\delta ^{2}n}\end{array} \), \(\begin{array}{l}\int_{0}^{1}f(x)x^{n}dx=0\end{array} \), \(\begin{array}{l}\int_{0}^{1}f(x)p(x)dx=0\end{array} \), \(\begin{array}{l}\int_{0}^{1}p_{n}f\rightarrow \int _{0}^{1}f^{2}\end{array} \), \(\begin{array}{l}\int_{0}^{1}p_{n}f = 0\end{array} \), \(\begin{array}{l}\int _{0}^{1}f^{2}=0\end{array} \), \(\begin{array}{l}\int_{0}^{1}f(x)dx = 0\end{array} \).
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