tanh Free Weierstrass Substitution Integration Calculator - integrate functions using the Weierstrass substitution method step by step at tan , rearranging, and taking the square roots yields. {\displaystyle t} How to type special characters on your Chromebook To enter a special unicode character using your Chromebook, type Ctrl + Shift + U. How do you get out of a corner when plotting yourself into a corner. Another way to get to the same point as C. Dubussy got to is the following: Click on a date/time to view the file as it appeared at that time. 0 Learn more about Stack Overflow the company, and our products. Learn more about Stack Overflow the company, and our products. The point. x where $\nu=x$ is $ab>0$ or $x+\pi$ if $ab<0$. Theorems on differentiation, continuity of differentiable functions. 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. + d : According to Spivak (2006, pp. Transfinity is the realm of numbers larger than every natural number: For every natural number k there are infinitely many natural numbers n > k. For a transfinite number t there is no natural number n t. We will first present the theory 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of t leads to following relationship between the inverse hyperbolic tangent cos Your Mobile number and Email id will not be published. 2 {\displaystyle t,} Weierstrass Substitution 24 4. Is there a way of solving integrals where the numerator is an integral of the denominator? 2 The trigonometric functions determine a function from angles to points on the unit circle, and by combining these two functions we have a function from angles to slopes. Weierstrass' preparation theorem. x cot {\textstyle t=-\cot {\frac {\psi }{2}}.}. / 2.3.8), which is an effective substitute for the Completeness Axiom, can easily be extended from sequences of numbers to sequences of points: Proposition 2.3.7 (Bolzano-Weierstrass Theorem). 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. Substitute methods had to be invented to . t preparation, we can state the Weierstrass Preparation Theorem, following [Krantz and Parks2002, Theorem 6.1.3]. Weierstrass Substitution and more integration techniques on https://brilliant.org/blackpenredpen/ This link gives you a 20% off discount on their annual prem. The integral on the left is $-\cot x$ and the one on the right is an easy $u$-sub with $u=\sin x$. One usual trick is the substitution $x=2y$. It only takes a minute to sign up. It yields: 8999. \(\Delta = -b_2^2 b_8 - 8b_4^3 - 27b_4^2 + 9b_2 b_4 b_6\). 1 rev2023.3.3.43278. Styling contours by colour and by line thickness in QGIS. It is based on the fact that trig. on the left hand side (and performing an appropriate variable substitution) {\displaystyle dt} So to get $\nu(t)$, you need to solve the integral Size of this PNG preview of this SVG file: 800 425 pixels. Why do we multiply numerator and denominator by $\sin px$ for evaluating $\int \frac{\cos ax+\cos bx}{1-2\cos cx}dx$? However, I can not find a decent or "simple" proof to follow. $$\ell=mr^2\frac{d\nu}{dt}=\text{constant}$$ Multivariable Calculus Review. From, This page was last modified on 15 February 2023, at 11:22 and is 2,352 bytes. "8. How can this new ban on drag possibly be considered constitutional? er. 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. Geometrically, the construction goes like this: for any point (cos , sin ) on the unit circle, draw the line passing through it and the point (1, 0). Alternatives for evaluating $ \int \frac { 1 } { 5 + 4 \cos x} \ dx $ ?? Why is there a voltage on my HDMI and coaxial cables? An affine transformation takes it to its Weierstrass form: If \(\mathrm{char} K \ne 2\) then we can further transform this to, \[Y^2 + a_1 XY + a_3 Y = X^3 + a_2 X^2 + a_4 X + a_6\]. These inequalities are two o f the most important inequalities in the supject of pro duct polynomials. Other sources refer to them merely as the half-angle formulas or half-angle formulae. where $\ell$ is the orbital angular momentum, $m$ is the mass of the orbiting body, the true anomaly $\nu$ is the angle in the orbit past periapsis, $t$ is the time, and $r$ is the distance to the attractor. [1] This follows since we have assumed 1 0 xnf (x) dx = 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). $$\sin E=\frac{\sqrt{1-e^2}\sin\nu}{1+e\cos\nu}$$ 2 For a special value = 1/8, we derive a . Geometrical and cinematic examples. ) Disconnect between goals and daily tasksIs it me, or the industry. Is it correct to use "the" before "materials used in making buildings are"? 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. For an even and $2\pi$ periodic function, why does $\int_{0}^{2\pi}f(x)dx = 2\int_{0}^{\pi}f(x)dx $. Tangent line to a function graph. Let f: [a,b] R be a real valued continuous 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). That is often appropriate when dealing with rational functions and with trigonometric functions. $\int \frac{dx}{\sin^3{x}}$ possible with universal substitution? Since [0, 1] is compact, the continuity of f implies uniform continuity. Modified 7 years, 6 months ago. 1 2 An irreducibe cubic with a flex can be affinely must be taken into account. rev2023.3.3.43278. Instead of Prohorov's theorem, we prove here a bare-hands substitute for the special case S = R. When doing so, it is convenient to have the following notion of convergence of distribution functions. 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. t &=-\frac{2}{1+u}+C \\ Of course it's a different story if $\left|\frac ba\right|\ge1$, where we get an unbound orbit, but that's a story for another bedtime. {\textstyle du=\left(-\csc x\cot x+\csc ^{2}x\right)\,dx} Proof. &=-\frac{2}{1+\text{tan}(x/2)}+C. \( 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). 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. $$\cos E=\frac{\cos\nu+e}{1+e\cos\nu}$$ It applies to trigonometric integrals that include a mixture of constants and trigonometric function. &=\text{ln}|u|-\frac{u^2}{2} + C \\ Definition 3.2.35. $$ Introducing a new variable My code is GPL licensed, can I issue a license to have my code be distributed in a specific MIT licensed project? 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). 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. the sum of the first n odds is n square proof by induction. = This is the discriminant. So you are integrating sum from 0 to infinity of (-1) n * t 2n / (2n+1) dt which is equal to the sum form 0 to infinity of (-1) n *t 2n+1 / (2n+1) 2 . p Let M = ||f|| exists as f is a continuous function on a compact set [0, 1]. , @robjohn : No, it's not "really the Weierstrass" since call the tangent half-angle substitution "the Weierstrass substitution" is incorrect. Now consider f is a continuous real-valued function on [0,1]. This is the \(j\)-invariant. A place where magic is studied and practiced? Splitting the numerator, and further simplifying: $\frac{1}{b}\int\frac{1}{\sin^2 x}dx-\frac{1}{b}\int\frac{\cos x}{\sin^2 x}dx=\frac{1}{b}\int\csc^2 x\:dx-\frac{1}{b}\int\frac{\cos x}{\sin^2 x}dx$. According to the theorem, every continuous function defined on a closed interval [a, b] can approximately be represented by a polynomial function. (originally defined for ) that is continuous but differentiable only on a set of points of measure zero. t 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. sin Redoing the align environment with a specific formatting. csc ) $$\begin{align}\int\frac{dx}{a+b\cos x}&=\frac1a\int\frac{d\nu}{1+e\cos\nu}=\frac12\frac1{\sqrt{1-e^2}}\int dE\\ Sie ist eine Variante der Integration durch Substitution, die auf bestimmte Integranden mit trigonometrischen Funktionen angewendet werden kann. By eliminating phi between the directly above and the initial definition of Vice versa, when a half-angle tangent is a rational number in the interval (0, 1) then the full-angle sine and cosine will both be rational, and there is a right triangle that has the full angle and that has side lengths that are a Pythagorean triple. By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. $$\int\frac{d\nu}{(1+e\cos\nu)^2}$$ q Integration of Some Other Classes of Functions 13", "Intgration des fonctions transcendentes", "19. t Integrate $\int \frac{\sin{2x}}{\sin{x}+\cos^2{x}}dx$, Find the indefinite integral $\int \frac{25}{(3\cos(x)+4\sin(x))^2} dx$. Benannt ist die Methode nach dem Mathematiker Karl Weierstra, der . &=\int{\frac{2du}{1+2u+u^2}} \\ Follow Up: struct sockaddr storage initialization by network format-string. artanh 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. As x varies, the point (cos x . Is it suspicious or odd to stand by the gate of a GA airport watching the planes? u Chain rule. {\textstyle \cos ^{2}{\tfrac {x}{2}},} arbor park school district 145 salary schedule; Tags . \implies Is there a single-word adjective for "having exceptionally strong moral principles"? {\textstyle t=\tan {\tfrac {x}{2}}} and then make the substitution of $t = \tan \frac{x}{2}$ in the integral. These two answers are the same because Site design / logo 2023 Stack Exchange Inc; user contributions licensed under CC BY-SA. for \(\mathrm{char} K \ne 2\), we have that if \((x,y)\) is a point, then \((x, -y)\) is Instead of + and , we have only one , at both ends of the real line. $\begingroup$ The name "Weierstrass substitution" is unfortunate, since Weierstrass didn't have anything to do with it (Stewart's calculus book to the contrary notwithstanding). cos 1 as follows: Using the double-angle formulas, introducing denominators equal to one thanks to the Pythagorean theorem, and then dividing numerators and denominators by {\textstyle t=\tanh {\tfrac {x}{2}}} 2 x What is the purpose of this D-shaped ring at the base of the tongue on my hiking boots? 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 . cos So as to relate the area swept out by a line segment joining the orbiting body to the attractor Kepler drew a little picture. That is, if. {\displaystyle t} It's not difficult to derive them using trigonometric identities. 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 \(-\pi\lt x\lt\pi\). {\displaystyle dx} Then we have. 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. We've added a "Necessary cookies only" option to the cookie consent popup, $\displaystyle\int_{0}^{2\pi}\frac{1}{a+ \cos\theta}\,d\theta$. One of the most important ways in which a metric is used is in approximation. {\textstyle t=0} Required fields are marked *, \(\begin{array}{l}\sum_{k=0}^{n}f\left ( \frac{k}{n} \right )\begin{pmatrix}n \\k\end{pmatrix}x_{k}(1-x)_{n-k}\end{array} \), \(\begin{array}{l}\sum_{k=0}^{n}(f-f(\zeta))\left ( \frac{k}{n} \right )\binom{n}{k} x^{k}(1-x)^{n-k}\end{array} \), \(\begin{array}{l}\sum_{k=0}^{n}\binom{n}{k}x^{k}(1-x)^{n-k} = (x+(1-x))^{n}=1\end{array} \), \(\begin{array}{l}\left|B_{n}(x, f)-f(\zeta) \right|=\left|B_{n}(x,f-f(\zeta)) \right|\end{array} \), \(\begin{array}{l}\leq B_{n}\left ( x,2M\left ( \frac{x- \zeta}{\delta } \right )^{2}+ \frac{\epsilon}{2} \right ) \end{array} \), \(\begin{array}{l}= \frac{2M}{\delta ^{2}} B_{n}(x,(x- \zeta )^{2})+ \frac{\epsilon}{2}\end{array} \), \(\begin{array}{l}B_{n}(x, (x- \zeta)^{2})= x^{2}+ \frac{1}{n}(x x^{2})-2 \zeta x + \zeta ^{2}\end{array} \), \(\begin{array}{l}\left| (B_{n}(x,f)-f(\zeta))\right|\leq \frac{\epsilon}{2}+\frac{2M}{\delta ^{2}}(x- \zeta)^{2}+\frac{2M}{\delta^{2}}\frac{1}{n}(x- x ^{2})\end{array} \), \(\begin{array}{l}\left| (B_{n}(x,f)-f(\zeta))\right|\leq \frac{\epsilon}{2}+\frac{2M}{\delta ^{2}}\frac{1}{n}(\zeta- \zeta ^{2})\end{array} \), \(\begin{array}{l}\left| (B_{n}(x,f)-f(\zeta))\right|\leq \frac{\epsilon}{2}+\frac{M}{2\delta ^{2}n}\end{array} \), \(\begin{array}{l}\int_{0}^{1}f(x)x^{n}dx=0\end{array} \), \(\begin{array}{l}\int_{0}^{1}f(x)p(x)dx=0\end{array} \), \(\begin{array}{l}\int_{0}^{1}p_{n}f\rightarrow \int _{0}^{1}f^{2}\end{array} \), \(\begin{array}{l}\int_{0}^{1}p_{n}f = 0\end{array} \), \(\begin{array}{l}\int _{0}^{1}f^{2}=0\end{array} \), \(\begin{array}{l}\int_{0}^{1}f(x)dx = 0\end{array} \). q According to Spivak (2006, pp. This is the one-dimensional stereographic projection of the unit circle . What is the correct way to screw wall and ceiling drywalls? Thus, the tangent half-angle formulae give conversions between the stereographic coordinate t on the unit circle and the standard angular coordinate . {\displaystyle t,} 2006, p.39). ( (2/2) The tangent half-angle substitution illustrated as stereographic projection of the circle. + Find reduction formulas for R x nex dx and R x sinxdx. This is really the Weierstrass substitution since $t=\tan(x/2)$. Trigonometric Substitution 25 5. The The steps for a proof by contradiction are: Step 1: Take the statement, and assume that the contrary is true (i.e. We use the universal trigonometric substitution: Since \(\sin x = {\frac{{2t}}{{1 + {t^2}}}},\) we have. 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." Irreducible cubics containing singular points can be affinely transformed by setting 2 6. Weierstrass's theorem has a far-reaching generalizationStone's theorem. x sin x Here we shall see the proof by using Bernstein Polynomial. gives, Taking the quotient of the formulae for sine and cosine yields. = f p < / M. We also know that 1 0 p(x)f (x) dx = 0. This is the one-dimensional stereographic projection of the unit circle parametrized by angle measure onto the real line. Published by at 29, 2022. Polynomial functions are simple functions that even computers can easily process, hence the Weierstrass Approximation theorem has great practical as well as theoretical utility. The sigma and zeta Weierstrass functions were introduced in the works of F . Preparation theorem. \begin{align} 2 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. The reason it is so powerful is that with Algebraic integrands you have numerous standard techniques for finding the AntiDerivative . B n (x, f) := \int{\frac{dx}{1+\text{sin}x}}&=\int{\frac{1}{1+2u/(1+u^{2})}\frac{2}{1+u^2}du} \\ Vol. Find $\int_0^{2\pi} \frac{1}{3 + \cos x} dx$. Remember that f and g are inverses of each other! 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 . \end{align} ) The Bolzano Weierstrass theorem is named after mathematicians Bernard Bolzano and Karl Weierstrass. However, I can not find a decent or "simple" proof to follow. Integration of rational functions by partial fractions 26 5.1. How do I align things in the following tabular environment? ( t 2.4: The Bolazno-Weierstrass Theorem - Mathematics LibreTexts International Symposium on History of Machines and Mechanisms. 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 . 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. dx&=\frac{2du}{1+u^2} This paper studies a perturbative approach for the double sine-Gordon equation. = [7] Michael Spivak called it the "world's sneakiest substitution".[8]. follows is sometimes called the Weierstrass substitution. The Bolzano-Weierstrass Theorem is at the foundation of many results in analysis. Basically it takes a rational trigonometric integrand and converts it to a rational algebraic integrand via substitutions. Benannt ist die Methode nach dem Mathematiker Karl Weierstra, der sie entwickelte. {\textstyle t=\tan {\tfrac {x}{2}}} 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 $$\frac12a^2(E-e\sin E)$$ 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. 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). Weierstrass, Karl (1915) [1875]. 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.