File usage on Commons. Integrate $\int \frac{4}{5+3\cos(2x)}\,d x$. Date/Time Thumbnail Dimensions User A line through P (except the vertical line) is determined by its slope. Thus, the tangent half-angle formulae give conversions between the stereographic coordinate t on the unit circle and the standard angular coordinate . [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. Styling contours by colour and by line thickness in QGIS. The reason it is so powerful is that with Algebraic integrands you have numerous standard techniques for finding the AntiDerivative . Did any DOS compatibility layers exist for any UNIX-like systems before DOS started to become outmoded? x x Step 2: Start an argument from the assumed statement and work it towards the conclusion.Step 3: While doing so, you should reach a contradiction.This means that this alternative statement is false, and thus we . csc {\textstyle x=\pi } The Weierstrass substitution is the trigonometric substitution which transforms an integral of the form. Especially, when it comes to polynomial interpolations in numerical analysis. These two answers are the same because As with other properties shared between the trigonometric functions and the hyperbolic functions, it is possible to use hyperbolic identities to construct a similar form of the substitution, \end{align*} Bibliography. As t goes from 1 to0, the point follows the part of the circle in the fourth quadrant from (0,1) to(1,0). Instead of + and , we have only one , at both ends of the real line. What can a lawyer do if the client wants him to be acquitted of everything despite serious evidence? 2 Here we shall see the proof by using Bernstein Polynomial. {\textstyle x} by the substitution The essence of this theorem is that no matter how much complicated the function f is given, we can always find a polynomial that is as close to f as we desire. . t Geometrical and cinematic examples. that is, |f(x) f()| 2M [(x )/ ]2 + /2 x [0, 1]. The Bolzano-Weierstrass Theorem says that no matter how " random " the sequence ( x n) may be, as long as it is bounded then some part of it must converge. The Bolzano-Weierstrass Property and Compactness. 2 Newton potential for Neumann problem on unit disk. From MathWorld--A Wolfram Web Resource. Finally, since t=tan(x2), solving for x yields that x=2arctant. 2 In integral calculus, the tangent half-angle substitution is a change of variables used for evaluating integrals, which converts a rational function of trigonometric functions of d 1 The Weierstrass substitution is an application of Integration by Substitution. One usual trick is the substitution $x=2y$. ) . {\textstyle t=\tan {\tfrac {x}{2}},} x For a special value = 1/8, we derive a . Polynomial functions are simple functions that even computers can easily process, hence the Weierstrass Approximation theorem has great practical as well as theoretical utility. Preparation theorem. |Algebra|. t [Reducible cubics consist of a line and a conic, which {\displaystyle 1+\tan ^{2}\alpha =1{\big /}\cos ^{2}\alpha } 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. arbor park school district 145 salary schedule; Tags . Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. [7] Michael Spivak called it the "world's sneakiest substitution".[8]. Evaluating $\int \frac{x\sin x-\cos x}{x\left(2\cos x+x-x\sin x\right)} {\rm d} x$ using elementary methods, Integrating $\int \frac{dx}{\sin^2 x \cos^2x-6\sin x\cos x}$. Finally, fifty years after Riemann, D. Hilbert . This entry briefly describes the history and significance of Alfred North Whitehead and Bertrand Russell's monumental but little read classic of symbolic logic, Principia Mathematica (PM), first published in 1910-1913. The plots above show for (red), 3 (green), and 4 (blue). How to solve this without using the Weierstrass substitution \[ \int . However, I can not find a decent or "simple" proof to follow. This proves the theorem for continuous functions on [0, 1]. How do you get out of a corner when plotting yourself into a corner. We can confirm the above result using a standard method of evaluating the cosecant integral by multiplying the numerator and denominator by sin = In the unit circle, application of the above shows that 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 Now consider f is a continuous real-valued function on [0,1]. brian kim, cpa clearvalue tax net worth . Follow Up: struct sockaddr storage initialization by network format-string. = 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. Tangent line to a function graph. The Weierstrass substitution is an application of Integration by Substitution . Why are physically impossible and logically impossible concepts considered separate in terms of probability? ) The Weierstrass representation is particularly useful for constructing immersed minimal surfaces. x ISBN978-1-4020-2203-6. $\int\frac{a-b\cos x}{(a^2-b^2)+b^2(\sin^2 x)}dx$. If \(a_1 = a_3 = 0\) (which is always the case 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). Apply for Mathematics with a Foundation Year - BSc (Hons) Undergraduate applications open for 2024 entry on 16 May 2023. In Ceccarelli, Marco (ed.). ) Redoing the align environment with a specific formatting. (c) Finally, use part b and the substitution y = f(x) to obtain the formula for R b a f(x)dx. into one of the form. Are there tables of wastage rates for different fruit and veg? "8. There are several ways of proving this theorem. A similar statement can be made about tanh /2. {\displaystyle a={\tfrac {1}{2}}(p+q)} \text{cos}x&=\frac{1-u^2}{1+u^2} \\ Follow Up: struct sockaddr storage initialization by network format-string, Linear Algebra - Linear transformation question. This equation can be further simplified through another affine transformation. Generalized version of the Weierstrass theorem. . Generally, if K is a subfield of the complex numbers then tan /2 K implies that {sin , cos , tan , sec , csc , cot } K {}. Now, let's return to the substitution formulas. Integrate $\int \frac{\sin{2x}}{\sin{x}+\cos^2{x}}dx$, Find the indefinite integral $\int \frac{25}{(3\cos(x)+4\sin(x))^2} dx$. Metadata. = 0 + 2\,\frac{dt}{1 + t^{2}} https://mathworld.wolfram.com/WeierstrassSubstitution.html. In the case = 0, we get the well-known perturbation theory for the sine-Gordon equation. cos t In the first line, one cannot simply substitute A place where magic is studied and practiced? 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. 3. Thus, Let N M/(22), then for n N, we have. Since, if 0 f Bn(x, f) and if g f Bn(x, f). a The tangent half-angle substitution in integral calculus, Learn how and when to remove this template message, https://en.wikipedia.org/w/index.php?title=Tangent_half-angle_formula&oldid=1119422059, This page was last edited on 1 November 2022, at 14:09. \(\Delta = -b_2^2 b_8 - 8b_4^3 - 27b_4^2 + 9b_2 b_4 b_6\). &=-\frac{2}{1+\text{tan}(x/2)}+C. How to handle a hobby that makes income in US. and the integral reads (1/2) The tangent half-angle substitution relates an angle to the slope of a line. ) Let E C ( X) be a closed subalgebra in C ( X ): 1 E . The tangent of half an angle is the stereographic projection of the circle onto a line. t derivatives are zero). If an integrand is a function of only \(\tan x,\) the substitution \(t = \tan x\) converts this integral into integral of a rational function. 2 2 eliminates the \(XY\) and \(Y\) terms. q . This follows since we have assumed 1 0 xnf (x) dx = 0 . 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. . {\textstyle t} 2 Let \(K\) denote the field we are working in. {\displaystyle \cos 2\alpha =\cos ^{2}\alpha -\sin ^{2}\alpha =1-2\sin ^{2}\alpha =2\cos ^{2}\alpha -1} 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. (d) Use what you have proven to evaluate R e 1 lnxdx. Brooks/Cole. or a singular point (a point where there is no tangent because both partial \). File usage on other wikis. The simplest proof I found is on chapter 3, "Why Does The Miracle Substitution Work?" The Bolzano Weierstrass theorem is named after mathematicians Bernard Bolzano and Karl Weierstrass. {\displaystyle b={\tfrac {1}{2}}(p-q)} It is also assumed that the reader is familiar with trigonometric and logarithmic identities. WEIERSTRASS APPROXIMATION THEOREM TL welll kroorn Neiendsaas . Is it known that BQP is not contained within NP? x 2 answers Score on last attempt: \( \quad 1 \) out of 3 Score in gradebook: 1 out of 3 At the beginning of 2000 , Miguel's house was worth 238 thousand dollars and Kyle's house was worth 126 thousand dollars. Viewed 270 times 2 $\begingroup$ After browsing some topics here, through one post, I discovered the "miraculous" Weierstrass substitutions. The proof of this theorem can be found in most elementary texts on real . Is there a way of solving integrals where the numerator is an integral of the denominator? 1 are well known as Weierstrass's inequality [1] or Weierstrass's Bernoulli's inequality [3]. ( 2011-01-12 01:01 Michael Hardy 927783 (7002 bytes) Illustration of the Weierstrass substitution, a parametrization of the circle used in integrating rational functions of sine and cosine. For a proof of Prohorov's theorem, which is beyond the scope of these notes, see [Dud89, Theorem 11.5.4]. $\qquad$. Moreover, since the partial sums are continuous (as nite sums of continuous functions), their uniform limit fis also continuous. transformed into a Weierstrass equation: We only consider cubic equations of this form. Proof by contradiction - key takeaways. Learn more about Stack Overflow the company, and our products. 2 The orbiting body has moved up to $Q^{\prime}$ at height So as to relate the area swept out by a line segment joining the orbiting body to the attractor Kepler drew a little picture. , Site design / logo 2023 Stack Exchange Inc; user contributions licensed under CC BY-SA. International Symposium on History of Machines and Mechanisms. 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. follows is sometimes called the Weierstrass substitution. the other point with the same \(x\)-coordinate. x How can Kepler know calculus before Newton/Leibniz were born ? Do new devs get fired if they can't solve a certain bug? The sigma and zeta Weierstrass functions were introduced in the works of F . cos 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 \(-\pi\lt x\lt\pi\). = Then substitute back that t=tan (x/2).I don't know how you would solve this problem without series, and given the original problem you could . = It applies to trigonometric integrals that include a mixture of constants and trigonometric function. The best answers are voted up and rise to the top, Not the answer you're looking for? Connect and share knowledge within a single location that is structured and easy to search. $$\int\frac{dx}{a+b\cos x}=\frac1a\int\frac{dx}{1+\frac ba\cos x}=\frac1a\int\frac{d\nu}{1+\left|\frac ba\right|\cos\nu}$$ \implies & d\theta = (2)'\!\cdot\arctan\left(t\right) + 2\!\cdot\!\big(\arctan\left(t\right)\big)' 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. t 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.. (This substitution is also known as the universal trigonometric substitution.) Size of this PNG preview of this SVG file: 800 425 pixels. We give a variant of the formulation of the theorem of Stone: Theorem 1. Weierstrass Substitution 24 4. doi:10.1145/174603.174409. \(j = c_4^3 / \Delta\) for \(\Delta \ne 0\). The Weierstrass Approximation theorem The Weierstrass Function Math 104 Proof of Theorem. Mathematische Werke von Karl Weierstrass (in German). Integration of Some Other Classes of Functions 13", "Intgration des fonctions transcendentes", "19. H We use the universal trigonometric substitution: Since \(\sin x = {\frac{{2t}}{{1 + {t^2}}}},\) we have. x File history. sin 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. and Transactions on Mathematical Software. (1) F(x) = R x2 1 tdt. \int{\frac{dx}{\text{sin}x+\text{tan}x}}&=\int{\frac{1}{\frac{2u}{1+u^2}+\frac{2u}{1-u^2}}\frac{2}{1+u^2}du} \\ . {\displaystyle t} (originally defined for ) that is continuous but differentiable only on a set of points of measure zero. The Gudermannian function gives a direct relationship between the circular functions and the hyperbolic ones that does not involve complex numbers. {\textstyle \cos ^{2}{\tfrac {x}{2}},} &=\int{\frac{2(1-u^{2})}{2u}du} \\ 2 cot 2 Modified 7 years, 6 months ago. tan \int{\frac{dx}{1+\text{sin}x}}&=\int{\frac{1}{1+2u/(1+u^{2})}\frac{2}{1+u^2}du} \\ Introducing a new variable Or, if you could kindly suggest other sources. = By eliminating phi between the directly above and the initial definition of |x y| |f(x) f(y)| /2 for every x, y [0, 1]. artanh $$\cos E=\frac{\cos\nu+e}{1+e\cos\nu}$$ For an even and $2\pi$ periodic function, why does $\int_{0}^{2\pi}f(x)dx = 2\int_{0}^{\pi}f(x)dx $. where gd() is the Gudermannian function. \). ) the \(X^2\) term (whereas if \(\mathrm{char} K = 3\) we can eliminate either the \(X^2\) 2 . 2.4: The Bolazno-Weierstrass Theorem - Mathematics LibreTexts To subscribe to this RSS feed, copy and paste this URL into your RSS reader. The Weierstrass substitution can also be useful in computing a Grbner basis to eliminate trigonometric functions from a . {\textstyle \int dx/(a+b\cos x)} Die Weierstra-Substitution (auch unter Halbwinkelmethode bekannt) ist eine Methode aus dem mathematischen Teilgebiet der Analysis. Some sources call these results the tangent-of-half-angle formulae . Free Weierstrass Substitution Integration Calculator - integrate functions using the Weierstrass substitution method step by step 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 \(x = 2\arctan t.\). 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. 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. Merlet, Jean-Pierre (2004). File:Weierstrass substitution.svg. Then we have. &= \frac{1}{(a - b) \sin^2 \frac{x}{2} + (a + b) \cos^2 \frac{x}{2}}\\ where $\nu=x$ is $ab>0$ or $x+\pi$ if $ab<0$. 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 \(t = \cos x.\). {\textstyle t=0} \begin{aligned} Connect and share knowledge within a single location that is structured and easy to search. What is the correct way to screw wall and ceiling drywalls? a \text{tan}x&=\frac{2u}{1-u^2} \\ 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.
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