weierstrass substitution proof

Other trigonometric functions can be written in terms of sine and cosine. The Weierstrass approximation theorem - University of St Andrews a ( The Weierstrass substitution is an application of Integration by Substitution. t The plots above show for (red), 3 (green), and 4 (blue). Styling contours by colour and by line thickness in QGIS. 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. for both limits of integration. PDF Introduction {\textstyle t=\tan {\tfrac {x}{2}},} A place where magic is studied and practiced? This is really the Weierstrass substitution since $t=\tan(x/2)$. [7] Michael Spivak called it the "world's sneakiest substitution".[8]. Some sources call these results the tangent-of-half-angle formulae . The Weierstrass Substitution (Introduction) | ExamSolutions {\textstyle x} According to the theorem, every continuous function defined on a closed interval [a, b] can approximately be represented by a polynomial function. Irreducible cubics containing singular points can be affinely transformed eliminates the \(XY\) and \(Y\) terms. The Weierstrass Substitution - Alexander Bogomolny 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. . [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. Tangent half-angle substitution - Wikipedia 1 sin Now we see that $e=\left|\frac ba\right|$, and we can use the eccentric anomaly, {\displaystyle b={\tfrac {1}{2}}(p-q)} $$. Proof by Contradiction (Maths): Definition & Examples - StudySmarter US Define: \(b_8 = a_1^2 a_6 + 4a_2 a_6 - a_1 a_3 a_4 + a_2 a_3^2 - a_4^2\). File usage on other wikis. File. The above descriptions of the tangent half-angle formulae (projection the unit circle and standard hyperbola onto the y-axis) give a geometric interpretation of this function. $\int\frac{a-b\cos x}{(a^2-b^2)+b^2(\sin^2 x)}dx$. All new items; Books; Journal articles; Manuscripts; Topics. Here you are shown the Weierstrass Substitution to help solve trigonometric integrals.Useful videos: Weierstrass Substitution continued: https://youtu.be/SkF. {\textstyle t=\tan {\tfrac {x}{2}}} Connect and share knowledge within a single location that is structured and easy to search. t can be expressed as the product of 2 + {\textstyle t} As x varies, the point (cos x . Weierstrass Substitution - Page 2 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. Hoelder functions. 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. by the substitution The Bolzano Weierstrass theorem is named after mathematicians Bernard Bolzano and Karl Weierstrass. If an integrand is a function of only \(\tan x,\) the substitution \(t = \tan x\) converts this integral into integral of a rational function. / in his 1768 integral calculus textbook,[3] and Adrien-Marie Legendre described the general method in 1817. 4 Parametrize each of the curves in R 3 described below a The $$y=\frac{a\sqrt{1-e^2}\sin\nu}{1+e\cos\nu}$$But still $$x=\frac{a(1-e^2)\cos\nu}{1+e\cos\nu}$$ Apply for Mathematics with a Foundation Year - BSc (Hons) Undergraduate applications open for 2024 entry on 16 May 2023. PDF Rationalizing Substitutions - Carleton Why do academics stay as adjuncts for years rather than move around? It applies to trigonometric integrals that include a mixture of constants and trigonometric function. x Metadata. Integration of Some Other Classes of Functions 13", "Intgration des fonctions transcendentes", "19. Modified 7 years, 6 months ago. Here we shall see the proof by using Bernstein Polynomial. by setting It is also assumed that the reader is familiar with trigonometric and logarithmic identities. Bernard Bolzano (Stanford Encyclopedia of Philosophy/Winter 2022 Edition) Weierstrass Function -- from Wolfram MathWorld The substitution - db0nus869y26v.cloudfront.net The reason it is so powerful is that with Algebraic integrands you have numerous standard techniques for finding the AntiDerivative . Theorems on differentiation, continuity of differentiable functions. into an ordinary rational function of Then the integral is written as. Thus there exists a polynomial p p such that f p </M. where $a$ and $e$ are the semimajor axis and eccentricity of the ellipse. are easy to study.]. |Algebra|. 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. G My code is GPL licensed, can I issue a license to have my code be distributed in a specific MIT licensed project? Proof Technique. = Tangent half-angle substitution - Wikiwand Finding $\int \frac{dx}{a+b \cos x}$ without Weierstrass substitution. 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). 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. Following this path, we are able to obtain a system of differential equations that shows the amplitude and phase modulation of the approximate solution. Date/Time Thumbnail Dimensions User t How to handle a hobby that makes income in US. Instead of + and , we have only one , at both ends of the real line. 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. Later authors, citing Stewart, have sometimes referred to this as the Weierstrass substitution, for instance: Jeffrey, David J.; Rich, Albert D. (1994). How can this new ban on drag possibly be considered constitutional? rev2023.3.3.43278. 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 $\int \frac{dx}{\sin^2 x \cos^2x-6\sin x\cos x}$. Did any DOS compatibility layers exist for any UNIX-like systems before DOS started to become outmoded? One can play an entirely analogous game with the hyperbolic functions. The formulation throughout was based on theta functions, and included much more information than this summary suggests. {\displaystyle t} 1 The Weierstrass substitution can also be useful in computing a Grbner basis to eliminate trigonometric functions from a . {\textstyle t=\tan {\tfrac {x}{2}}} Published by at 29, 2022. &=\frac1a\frac1{\sqrt{1-e^2}}E+C=\frac{\text{sgn}\,a}{\sqrt{a^2-b^2}}\sin^{-1}\left(\frac{\sqrt{a^2-b^2}\sin\nu}{|a|+|b|\cos\nu}\right)+C\\&=\frac{1}{\sqrt{a^2-b^2}}\sin^{-1}\left(\frac{\sqrt{a^2-b^2}\sin x}{a+b\cos x}\right)+C\end{align}$$ 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). Describe where the following function is di erentiable and com-pute its derivative. It only takes a minute to sign up. Fact: Isomorphic curves over some field \(K\) have the same \(j\)-invariant. If $a=b$ then you can modify the technique for $a=b=1$ slightly to obtain: $\int \frac{dx}{b+b\cos x}=\int\frac{b-b\cos x}{(b+b\cos x)(b-b\cos x)}dx$, $=\int\frac{b-b\cos x}{b^2-b^2\cos^2 x}dx=\int\frac{b-b\cos x}{b^2(1-\cos^2 x)}dx=\frac{1}{b}\int\frac{1-\cos x}{\sin^2 x}dx$. In other words, if f is a continuous real-valued function on [a, b] and if any > 0 is given, then there exist a polynomial P on [a, b] such that |f(x) P(x)| < , for every x in [a, b]. , differentiation rules imply. Did any DOS compatibility layers exist for any UNIX-like systems before DOS started to become outmoded? We use the universal trigonometric substitution: Since \(\sin x = {\frac{{2t}}{{1 + {t^2}}}},\) we have. , rearranging, and taking the square roots yields. If we identify the parameter t in both cases we arrive at a relationship between the circular functions and the hyperbolic ones. &=\int{(\frac{1}{u}-u)du} \\ 2 An affine transformation takes it to its Weierstrass form: If \(\mathrm{char} K \ne 2\) then we can further transform this to, \[Y^2 + a_1 XY + a_3 Y = X^3 + a_2 X^2 + a_4 X + a_6\]. The Weierstrass substitution formulas for -Weierstrass substitution | Physics Forums S2CID13891212. Why do academics stay as adjuncts for years rather than move around? weierstrass substitution proof He is best known for the Casorati Weierstrass theorem in complex analysis. "Weierstrass Substitution". Tangent line to a function graph. Note that these are just the formulas involving radicals (http://planetmath.org/Radical6) as designated in the entry goniometric formulas; however, due to the restriction on x, the s are unnecessary. 2 x ) and = 0 + 2\,\frac{dt}{1 + t^{2}} doi:10.1007/1-4020-2204-2_16. . 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 . Especially, when it comes to polynomial interpolations in numerical analysis. 20 (1): 124135. x To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Then Kepler's first law, the law of trajectory, is What is a word for the arcane equivalent of a monastery? It is just the Chain Rule, written in terms of integration via the undamenFtal Theorem of Calculus. t = \tan \left(\frac{\theta}{2}\right) \implies 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). ( 7.3: The Bolzano-Weierstrass Theorem - Mathematics LibreTexts File:Weierstrass substitution.svg. PDF Ects: 8 $\qquad$. preparation, we can state the Weierstrass Preparation Theorem, following [Krantz and Parks2002, Theorem 6.1.3]. cos 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. \\ or a singular point (a point where there is no tangent because both partial The Weierstrass substitution is the trigonometric substitution which transforms an integral of the form. 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 \(t = \cos x.\). Michael Spivak escreveu que "A substituio mais . Weierstrass Trig Substitution Proof - Mathematics Stack Exchange {\displaystyle dx} \begin{align} ( Split the numerator again, and use pythagorean identity. Remember that f and g are inverses of each other! Is it known that BQP is not contained within NP? Kluwer. If you do use this by t the power goes to 2n. 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, What is the correct way to screw wall and ceiling drywalls? 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). 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 Differentiation: Derivative of a real function. We can confirm the above result using a standard method of evaluating the cosecant integral by multiplying the numerator and denominator by , 4. Weisstein, Eric W. (2011). "1.4.6. He also derived a short elementary proof of Stone Weierstrass theorem. Weierstrass Approximation theorem provides an important result of approximating a given continuous function defined on a closed interval to a polynomial function, which can be easily computed to find the value of the function. 1 $$\cos E=\frac{\cos\nu+e}{1+e\cos\nu}$$ Mathematica GuideBook for Symbolics. {\displaystyle t=\tan {\tfrac {1}{2}}\varphi } PDF Calculus MATH 172-Fall 2017 Lecture Notes - Texas A&M University How to handle a hobby that makes income in US, Trying to understand how to get this basic Fourier Series. The For a special value = 1/8, we derive a . 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 ). We have a rational expression in and in the denominator, so we use the Weierstrass substitution to simplify the integral: and. Elliptic functions with critical orbits approaching infinity \int{\frac{dx}{1+\text{sin}x}}&=\int{\frac{1}{1+2u/(1+u^{2})}\frac{2}{1+u^2}du} \\ and a rational function of The steps for a proof by contradiction are: Step 1: Take the statement, and assume that the contrary is true (i.e. 195200. Are there tables of wastage rates for different fruit and veg? The technique of Weierstrass Substitution is also known as tangent half-angle substitution . weierstrass substitution proof. Site design / logo 2023 Stack Exchange Inc; user contributions licensed under CC BY-SA. Required fields are marked *, \(\begin{array}{l}\sum_{k=0}^{n}f\left ( \frac{k}{n} \right )\begin{pmatrix}n \\k\end{pmatrix}x_{k}(1-x)_{n-k}\end{array} \), \(\begin{array}{l}\sum_{k=0}^{n}(f-f(\zeta))\left ( \frac{k}{n} \right )\binom{n}{k} x^{k}(1-x)^{n-k}\end{array} \), \(\begin{array}{l}\sum_{k=0}^{n}\binom{n}{k}x^{k}(1-x)^{n-k} = (x+(1-x))^{n}=1\end{array} \), \(\begin{array}{l}\left|B_{n}(x, f)-f(\zeta) \right|=\left|B_{n}(x,f-f(\zeta)) \right|\end{array} \), \(\begin{array}{l}\leq B_{n}\left ( x,2M\left ( \frac{x- \zeta}{\delta } \right )^{2}+ \frac{\epsilon}{2} \right ) \end{array} \), \(\begin{array}{l}= \frac{2M}{\delta ^{2}} B_{n}(x,(x- \zeta )^{2})+ \frac{\epsilon}{2}\end{array} \), \(\begin{array}{l}B_{n}(x, (x- \zeta)^{2})= x^{2}+ \frac{1}{n}(x x^{2})-2 \zeta x + \zeta ^{2}\end{array} \), \(\begin{array}{l}\left| (B_{n}(x,f)-f(\zeta))\right|\leq \frac{\epsilon}{2}+\frac{2M}{\delta ^{2}}(x- \zeta)^{2}+\frac{2M}{\delta^{2}}\frac{1}{n}(x- x ^{2})\end{array} \), \(\begin{array}{l}\left| (B_{n}(x,f)-f(\zeta))\right|\leq \frac{\epsilon}{2}+\frac{2M}{\delta ^{2}}\frac{1}{n}(\zeta- \zeta ^{2})\end{array} \), \(\begin{array}{l}\left| (B_{n}(x,f)-f(\zeta))\right|\leq \frac{\epsilon}{2}+\frac{M}{2\delta ^{2}n}\end{array} \), \(\begin{array}{l}\int_{0}^{1}f(x)x^{n}dx=0\end{array} \), \(\begin{array}{l}\int_{0}^{1}f(x)p(x)dx=0\end{array} \), \(\begin{array}{l}\int_{0}^{1}p_{n}f\rightarrow \int _{0}^{1}f^{2}\end{array} \), \(\begin{array}{l}\int_{0}^{1}p_{n}f = 0\end{array} \), \(\begin{array}{l}\int _{0}^{1}f^{2}=0\end{array} \), \(\begin{array}{l}\int_{0}^{1}f(x)dx = 0\end{array} \). According to Spivak (2006, pp. t Weierstrass's theorem has a far-reaching generalizationStone's theorem. However, I can not find a decent or "simple" proof to follow. Merlet, Jean-Pierre (2004). cos 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. Integrating $I=\int^{\pi}_0\frac{x}{1-\cos{\beta}\sin{x}}dx$ without Weierstrass Substitution. Ask Question Asked 7 years, 9 months ago. That is often appropriate when dealing with rational functions and with trigonometric functions. There are several ways of proving this theorem. To calculate an integral of the form \(\int {R\left( {\sin x} \right)\cos x\,dx} ,\) where both functions \(\sin x\) and \(\cos x\) have even powers, use the substitution \(t = \tan x\) and the formulas. Free Weierstrass Substitution Integration Calculator - integrate functions using the Weierstrass substitution method step by step Is there a proper earth ground point in this switch box? tan . Karl Weierstrass | German mathematician | Britannica Integrate $\int \frac{\sin{2x}}{\sin{x}+\cos^2{x}}dx$, Find the indefinite integral $\int \frac{25}{(3\cos(x)+4\sin(x))^2} dx$. You can still apply for courses starting in 2023 via the UCAS website. + He gave this result when he was 70 years old. 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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