Some sources call these results the tangent-of-half-angle formulae. of its coperiodic Weierstrass function and in terms of associated Jacobian functions; he also located its poles and gave expressions for its fundamental periods. You can still apply for courses starting in 2023 via the UCAS website. cos = The tangent half-angle substitution parametrizes the unit circle centered at (0, 0). Ask Question Asked 7 years, 9 months ago. 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. csc cos d cot File. {\displaystyle dx} 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. The substitution is: u tan 2. for < < , u R . Definition 3.2.35. |Contact| Fact: The discriminant is zero if and only if the curve is singular. t The plots above show for (red), 3 (green), and 4 (blue). NCERT Solutions Class 12 Business Studies, NCERT Solutions Class 12 Accountancy Part 1, NCERT Solutions Class 12 Accountancy Part 2, NCERT Solutions Class 11 Business Studies, NCERT Solutions for Class 10 Social Science, NCERT Solutions for Class 10 Maths Chapter 1, NCERT Solutions for Class 10 Maths Chapter 2, NCERT Solutions for Class 10 Maths Chapter 3, NCERT Solutions for Class 10 Maths Chapter 4, NCERT Solutions for Class 10 Maths Chapter 5, NCERT Solutions for Class 10 Maths Chapter 6, NCERT Solutions for Class 10 Maths Chapter 7, NCERT Solutions for Class 10 Maths Chapter 8, NCERT Solutions for Class 10 Maths Chapter 9, NCERT Solutions for Class 10 Maths Chapter 10, NCERT Solutions for Class 10 Maths Chapter 11, NCERT Solutions for Class 10 Maths Chapter 12, NCERT Solutions for Class 10 Maths Chapter 13, NCERT Solutions for Class 10 Maths Chapter 14, NCERT Solutions for Class 10 Maths Chapter 15, NCERT Solutions for Class 10 Science Chapter 1, NCERT Solutions for Class 10 Science Chapter 2, NCERT Solutions for Class 10 Science Chapter 3, NCERT Solutions for Class 10 Science Chapter 4, NCERT Solutions for Class 10 Science Chapter 5, NCERT Solutions for Class 10 Science Chapter 6, NCERT Solutions for Class 10 Science Chapter 7, NCERT Solutions for Class 10 Science Chapter 8, NCERT Solutions for Class 10 Science Chapter 9, NCERT Solutions for Class 10 Science Chapter 10, NCERT Solutions for Class 10 Science Chapter 11, NCERT Solutions for Class 10 Science Chapter 12, NCERT Solutions for Class 10 Science Chapter 13, NCERT Solutions for Class 10 Science Chapter 14, NCERT Solutions for Class 10 Science Chapter 15, NCERT Solutions for Class 10 Science Chapter 16, NCERT Solutions For Class 9 Social Science, NCERT Solutions For Class 9 Maths Chapter 1, NCERT Solutions For Class 9 Maths Chapter 2, NCERT Solutions For Class 9 Maths Chapter 3, NCERT Solutions For Class 9 Maths Chapter 4, NCERT Solutions For Class 9 Maths Chapter 5, NCERT Solutions For Class 9 Maths Chapter 6, NCERT Solutions For Class 9 Maths Chapter 7, NCERT Solutions For Class 9 Maths Chapter 8, NCERT Solutions For Class 9 Maths Chapter 9, NCERT Solutions For Class 9 Maths Chapter 10, NCERT Solutions For Class 9 Maths Chapter 11, NCERT Solutions For Class 9 Maths Chapter 12, NCERT Solutions For Class 9 Maths Chapter 13, NCERT Solutions For Class 9 Maths Chapter 14, NCERT Solutions For Class 9 Maths Chapter 15, NCERT Solutions for Class 9 Science Chapter 1, NCERT Solutions for Class 9 Science Chapter 2, NCERT Solutions for Class 9 Science Chapter 3, NCERT Solutions for Class 9 Science Chapter 4, NCERT Solutions for Class 9 Science Chapter 5, NCERT Solutions for Class 9 Science Chapter 6, NCERT Solutions for Class 9 Science Chapter 7, NCERT Solutions for Class 9 Science Chapter 8, NCERT Solutions for Class 9 Science Chapter 9, NCERT Solutions for Class 9 Science Chapter 10, NCERT Solutions for Class 9 Science Chapter 11, NCERT Solutions for Class 9 Science Chapter 12, NCERT Solutions for Class 9 Science Chapter 13, NCERT Solutions for Class 9 Science Chapter 14, NCERT Solutions for Class 9 Science Chapter 15, NCERT Solutions for Class 8 Social Science, NCERT Solutions for Class 7 Social Science, NCERT Solutions For Class 6 Social Science, CBSE Previous Year Question Papers Class 10, CBSE Previous Year Question Papers Class 12, Linear Equations In Two Variables Class 9 Notes, Important Questions Class 8 Maths Chapter 4 Practical Geometry, CBSE Previous Year Question Papers Class 12 Maths, CBSE Previous Year Question Papers Class 10 Maths, ICSE Previous Year Question Papers Class 10, ISC Previous Year Question Papers Class 12 Maths, JEE Main 2023 Question Papers with Answers, JEE Main 2022 Question Papers with Answers, JEE Advanced 2022 Question Paper with Answers. 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. 195200. What can a lawyer do if the client wants him to be acquitted of everything despite serious evidence? x Weierstrass Approximation theorem in real analysis presents the notion of approximating continuous functions by polynomial functions. = My question is, from that chapter, can someone please explain to me how algebraically the $\frac{\theta}{2}$ angle is derived? How to integrate $\int \frac{\cos x}{1+a\cos x}\ dx$? tan {\textstyle x} [Reducible cubics consist of a line and a conic, which = by the substitution {\displaystyle b={\tfrac {1}{2}}(p-q)} \begin{align} By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. sines and cosines can be expressed as rational functions of A geometric proof of the Weierstrass substitution 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 t {\displaystyle t} . 2 One of the most important ways in which a metric is used is in approximation. or the \(X\) term). It applies to trigonometric integrals that include a mixture of constants and trigonometric function. He is best known for the Casorati Weierstrass theorem in complex analysis. in his 1768 integral calculus textbook,[3] and Adrien-Marie Legendre described the general method in 1817. \end{align} The general statement is something to the eect that Any rational function of sinx and cosx can be integrated using the . 6. tan d 1 The method is known as the Weierstrass substitution. t two values that \(Y\) may take. As x varies, the point (cosx,sinx) winds repeatedly around the unit circle centered at(0,0). t The editors were, apart from Jan Berg and Eduard Winter, Friedrich Kambartel, Jaromir Loul, Edgar Morscher and . q Site design / logo 2023 Stack Exchange Inc; user contributions licensed under CC BY-SA. preparation, we can state the Weierstrass Preparation Theorem, following [Krantz and Parks2002, Theorem 6.1.3]. b 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. MathWorld. cos \end{align} Elementary functions and their derivatives. In Weierstrass form, we see that for any given value of \(X\), there are at most 2 and the integral reads $$y=\frac{a\sqrt{1-e^2}\sin\nu}{1+e\cos\nu}$$But still $$x=\frac{a(1-e^2)\cos\nu}{1+e\cos\nu}$$ $$r=\frac{a(1-e^2)}{1+e\cos\nu}$$ Generally, if K is a subfield of the complex numbers then tan /2 K implies that {sin , cos , tan , sec , csc , cot } K {}. The best answers are voted up and rise to the top, Not the answer you're looking for? a pp. {\textstyle \csc x-\cot x=\tan {\tfrac {x}{2}}\colon }. Remember that f and g are inverses of each other! Metadata. 2 The Weierstrass substitution can also be useful in computing a Grbner basis to eliminate trigonometric functions from a system of equations (Trott {\textstyle t=\tan {\tfrac {x}{2}}} ( The Bolzano-Weierstrass Theorem is at the foundation of many results in analysis. x Chain rule. This approach was generalized by Karl Weierstrass to the Lindemann Weierstrass theorem. 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. Stewart provided no evidence for the attribution to Weierstrass. 3. t are well known as Weierstrass's inequality [1] or Weierstrass's Bernoulli's inequality [3]. Click or tap a problem to see the solution. How can Kepler know calculus before Newton/Leibniz were born ? 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. If tan /2 is a rational number then each of sin , cos , tan , sec , csc , and cot will be a rational number (or be infinite). \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} \\ Let M = ||f|| exists as f is a continuous function on a compact set [0, 1]. A line through P (except the vertical line) is determined by its slope. 2 Weierstrass Trig Substitution Proof. Two curves with the same \(j\)-invariant are isomorphic over \(\bar {K}\). CHANGE OF VARIABLE OR THE SUBSTITUTION RULE 7 Why do academics stay as adjuncts for years rather than move around? can be expressed as the product of t , , Proof of Weierstrass Approximation Theorem . \(j = c_4^3 / \Delta\) for \(\Delta \ne 0\). 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. How do you get out of a corner when plotting yourself into a corner. csc Weierstrass Substitution 24 4. It's not difficult to derive them using trigonometric identities. 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. = cosx=cos2(x2)-sin2(x2)=(11+t2)2-(t1+t2)2=11+t2-t21+t2=1-t21+t2. The Weierstrass substitution parametrizes the unit circle centered at (0, 0). Vol. ) In the first line, one cannot simply substitute The Bolzano-Weierstrass Property and Compactness. {\textstyle \int dx/(a+b\cos x)} rev2023.3.3.43278. 1 All Categories; Metaphysics and Epistemology From, This page was last modified on 15 February 2023, at 11:22 and is 2,352 bytes. of this paper: http://www.westga.edu/~faucette/research/Miracle.pdf. = = This is the one-dimensional stereographic projection of the unit circle . It only takes a minute to sign up. Then we can find polynomials pn(x) such that every pn converges uniformly to x on [a,b]. Trigonometric Substitution 25 5. James Stewart wasn't any good at history. \implies & d\theta = (2)'\!\cdot\arctan\left(t\right) + 2\!\cdot\!\big(\arctan\left(t\right)\big)' cot The point. Mayer & Mller. 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 \[\int {\frac{{dx}}{{\sec x + 1}}}.\]. Other trigonometric functions can be written in terms of sine and cosine. 2 eliminates the \(XY\) and \(Y\) terms. In the original integer, Combining the Pythagorean identity with the double-angle formula for the cosine, x Why are physically impossible and logically impossible concepts considered separate in terms of probability? 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. &=\text{ln}|\text{tan}(x/2)|-\frac{\text{tan}^2(x/2)}{2} + C. a We've added a "Necessary cookies only" option to the cookie consent popup, $\displaystyle\int_{0}^{2\pi}\frac{1}{a+ \cos\theta}\,d\theta$. This allows us to write the latter as rational functions of t (solutions are given below). Learn more about Stack Overflow the company, and our products. t ) 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. &=-\frac{2}{1+u}+C \\ A related substitution appears in Weierstrasss Mathematical Works, from an 1875 lecture wherein Weierstrass credits Carl Gauss (1818) with the idea of solving an integral of the form {\textstyle t=0} 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 . Or, if you could kindly suggest other sources. brian kim, cpa clearvalue tax net worth . H Thus there exists a polynomial p p such that f p </M. artanh Weierstrass' preparation theorem. Multivariable Calculus Review. if \(\mathrm{char} K \ne 3\), then a similar trick eliminates Connect and share knowledge within a single location that is structured and easy to search. x 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. Alternatives for evaluating $ \int \frac { 1 } { 5 + 4 \cos x} \ dx $ ?? Michael Spivak escreveu que "A substituio mais . 2.4: The Bolazno-Weierstrass Theorem - Mathematics LibreTexts 2 The Weierstrass substitution is very useful for integrals involving a simple rational expression in \(\sin x\) and/or \(\cos x\) in the denominator. 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 . Example 15. Is there a single-word adjective for "having exceptionally strong moral principles"? {\displaystyle t,} Likewise if tanh /2 is a rational number then each of sinh , cosh , tanh , sech , csch , and coth will be a rational number (or be infinite). From MathWorld--A Wolfram Web Resource. We have a rational expression in and in the denominator, so we use the Weierstrass substitution to simplify the integral: and. f p < / M. We also know that 1 0 p(x)f (x) dx = 0. 2 x p \( Following this path, we are able to obtain a system of differential equations that shows the amplitude and phase modulation of the approximate solution. csc = {\textstyle du=\left(-\csc x\cot x+\csc ^{2}x\right)\,dx} Given a function f, finding a sequence which converges to f in the metric d is called uniform approximation.The most important result in this area is due to the German mathematician Karl Weierstrass (1815 to 1897).. This equation can be further simplified through another affine transformation. There are several ways of proving this theorem. d Especially, when it comes to polynomial interpolations in numerical analysis. 1 Polynomial functions are simple functions that even computers can easily process, hence the Weierstrass Approximation theorem has great practical as well as theoretical utility. Your Mobile number and Email id will not be published. An irreducibe cubic with a flex can be affinely Follow Up: struct sockaddr storage initialization by network format-string. weierstrass substitution proof. 5. = Now, fix [0, 1]. Then the integral is written as. Did any DOS compatibility layers exist for any UNIX-like systems before DOS started to become outmoded? How can this new ban on drag possibly be considered constitutional? Benannt ist die Methode nach dem Mathematiker Karl Weierstra, der . {\textstyle u=\csc x-\cot x,} \implies weierstrass substitution proof. $\qquad$ $\endgroup$ - Michael Hardy the sum of the first n odds is n square proof by induction. Finding $\int \frac{dx}{a+b \cos x}$ without Weierstrass substitution. These identities are known collectively as the tangent half-angle formulae because of the definition of "8. Mathematische Werke von Karl Weierstrass (in German). \). After browsing some topics here, through one post, I discovered the "miraculous" Weierstrass substitutions. For an even and $2\pi$ periodic function, why does $\int_{0}^{2\pi}f(x)dx = 2\int_{0}^{\pi}f(x)dx $. By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Every bounded sequence of points in R 3 has a convergent subsequence. Finally, since t=tan(x2), solving for x yields that x=2arctant. $$\sin E=\frac{\sqrt{1-e^2}\sin\nu}{1+e\cos\nu}$$ Sie ist eine Variante der Integration durch Substitution, die auf bestimmte Integranden mit trigonometrischen Funktionen angewendet werden kann. This is the \(j\)-invariant. Preparation theorem. + 193. : The Weierstrass Approximation theorem is named after German mathematician Karl Theodor Wilhelm Weierstrass. B n (x, f) := identities (see Appendix C and the text) can be used to simplify such rational expressions once we make a preliminary substitution. b $$\int\frac{d\nu}{(1+e\cos\nu)^2}$$ / Complex Analysis - Exam. cos Brooks/Cole. Let f: [a,b] R be a real valued continuous function. Irreducible cubics containing singular points can be affinely transformed Theorems on differentiation, continuity of differentiable functions. . 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. 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 x {\\textstyle x} into an ordinary rational function of t {\\textstyle t} by setting t = tan x 2 {\\textstyle t=\\tan {\\tfrac {x}{2}}} . Integrate $\int \frac{\sin{2x}}{\sin{x}+\cos^2{x}}dx$, Find the indefinite integral $\int \frac{25}{(3\cos(x)+4\sin(x))^2} dx$. The general[1] transformation formula is: The tangent of half an angle is important in spherical trigonometry and was sometimes known in the 17th century as the half tangent or semi-tangent. The Weierstrass substitution can also be useful in computing a Grbner basis to eliminate trigonometric functions from a . His domineering father sent him to the University of Bonn at age 19 to study law and finance in preparation for a position in the Prussian civil service. 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. 20 (1): 124135. Example 3. 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 ). where gd() is the Gudermannian function. 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.