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Tangent half-angle substitution

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Change of variable for integrals involving trigonometric functions

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Thetangent half-angle substitution is achange of variables used for evaluatingintegrals, which converts arational function oftrigonometric functions of ${\textstyle x}$ into an ordinary rational function of ${\textstyle t}$ by setting ${\textstyle t=\tan {\tfrac {x}{2}}}$ . This is the one-dimensionalstereographic projection of theunit circle parametrized byangle measure onto thereal line. The general^[1] transformation formula is:

$\int f(\sin x,\cos x)\,dx=\int f{\left({\frac {2t}{1+t^{2}}},{\frac {1-t^{2}}{1+t^{2}}}\right)}{\frac {2\,dt}{1+t^{2}}}.$

The tangent of half an angle is important inspherical trigonometry and was sometimes known in the 17th century as the half tangent or semi-tangent.^[2]Leonhard Euler used it to evaluate the integral ${\textstyle \int dx/(a+b\cos x)}$ in his1768 integral calculus textbook,^[3] andAdrien-Marie Legendre described the general method in 1817.^[4]

The substitution is described in most integral calculus textbooks since the late 19th century, usually without any special name.^[5] It is known in Russia as theuniversal trigonometric substitution,^[6] and also known by variant names such ashalf-tangent substitution orhalf-angle substitution. It is sometimes misattributed as theWeierstrass substitution.^[7]Michael Spivak called it the "world's sneakiest substitution".^[8]

The substitution

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The tangent half-angle substitution relates an angle to the slope of a line.

Introducing a new variable ${\textstyle t=\tan {\tfrac {x}{2}},}$ sines and cosines can be expressed asrational functions of $t, {\displaystyle t,}$ and $d x {\displaystyle dx}$ can be expressed as the product of $d t {\displaystyle dt}$ and a rational function of $t, {\displaystyle t,}$ as follows: $\sin x={\frac {2t}{1+t^{2}}},\quad \cos x={\frac {1-t^{2}}{1+t^{2}}},\quad {\text{and}}\quad dx={\frac {2}{1+t^{2}}}\,dt.$

Similar expressions can be written fortanx,cotx,secx, andcscx.

Derivation

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Using thedouble-angle formulas $\sin x=2\sin {\tfrac {x}{2}}\cos {\tfrac {x}{2}}$ and $\cos x=\cos ^{2}{\tfrac {x}{2}}-\sin ^{2}{\tfrac {x}{2}}$ and introducing denominators equal to one by thePythagorean identity $1=\cos ^{2}{\tfrac {x}{2}}+\sin ^{2}{\tfrac {x}{2}}$ results in

${\begin{aligned}\sin x&={\frac {2\sin {\tfrac {x}{2}}\,\cos {\tfrac {x}{2}}}{\cos ^{2}{\tfrac {x}{2}}+\sin ^{2}{\tfrac {x}{2}}}}={\frac {2\tan {\tfrac {x}{2}}}{1+\tan ^{2}{\tfrac {x}{2}}}}={\frac {2t}{1+t^{2}}},\\[18mu]\cos x&={\frac {\cos ^{2}{\tfrac {x}{2}}-\sin ^{2}{\tfrac {x}{2}}}{\cos ^{2}{\tfrac {x}{2}}+\sin ^{2}{\tfrac {x}{2}}}}={\frac {1-\tan ^{2}{\tfrac {x}{2}}}{1+\tan ^{2}{\tfrac {x}{2}}}}={\frac {1-t^{2}}{1+t^{2}}}.\end{aligned}}$

Finally, since ${\textstyle t=\tan {\tfrac {x}{2}}}$ ,differentiation rules imply

$dt={\tfrac {1}{2}}\left(1+\tan ^{2}{\tfrac {x}{2}}\right)dx={\frac {1+t^{2}}{2}}\,dx,$ and thus $dx={\frac {2}{1+t^{2}}}\,dt.$

Examples

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Antiderivative of cosecant

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${\begin{aligned}\int \csc x\,dx&=\int {\frac {dx}{\sin x}}\\[6pt]&=\int \left({\frac {1+t^{2}}{2t}}\right)\left({\frac {2}{1+t^{2}}}\right)dt&&t=\tan {\tfrac {x}{2}}\\[6pt]&=\int {\frac {dt}{t}}\\[6pt]&=\ln |t|+C\\[6pt]&=\ln \left|\tan {\tfrac {x}{2}}\right|+C.\end{aligned}}$

We can confirm the above result using a standard method of evaluating the cosecant integral by multiplying the numerator and denominator by ${\textstyle \csc x-\cot x}$ and performing the substitution ${\textstyle u=\csc x-\cot x,}$ ${\textstyle du=\left(-\csc x\cot x+\csc ^{2}x\right)\,dx}$ . ${\begin{aligned}\int \csc x\,dx&=\int {\frac {\csc x(\csc x-\cot x)}{\csc x-\cot x}}\,dx\\[6pt]&=\int {\frac {\left(\csc ^{2}x-\csc x\cot x\right)\,dx}{\csc x-\cot x}}\qquad u=\csc x-\cot x\\[6pt]&=\int {\frac {du}{u}}\\[6pt]&=\ln |u|+C\\[6pt]&=\ln \left|\csc x-\cot x\right|+C.\end{aligned}}$

These two answers are the same because ${\textstyle \csc x-\cot x=\tan {\tfrac {x}{2}}\colon }$

${\begin{aligned}\csc x-\cot x&={\frac {1}{\sin x}}-{\frac {\cos x}{\sin x}}\\[6pt]&={\frac {1+t^{2}}{2t}}-{\frac {1-t^{2}}{1+t^{2}}}{\frac {1+t^{2}}{2t}}\qquad \qquad t=\tan {\tfrac {x}{2}}\\[6pt]&={\frac {2t^{2}}{2t}}=t\\[6pt]&=\tan {\tfrac {x}{2}}\end{aligned}}$

Thesecant integral may be evaluated in a similar manner.

A definite integral

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We wish to evaluate the integral:

$\int _{0}^{2\pi }{\frac {dx}{2+\cos x}}$

A naïve approach splits the interval and applies the substitution $t=\tan {\frac {x}{2}}$ . However, this substitution has a singularity at $x=\pi$ , which corresponds to a vertical asymptote. Therefore, the integral must be split at that point and handled carefully:

${\begin{aligned}\int _{0}^{2\pi }{\frac {dx}{2+\cos x}}&=\int _{0}^{\pi }{\frac {dx}{2+\cos x}}+\int _{\pi }^{2\pi }{\frac {dx}{2+\cos x}}\\[6pt]&=\int _{0}^{\infty }{\frac {2\,dt}{3+t^{2}}}+\int _{-\infty }^{0}{\frac {2\,dt}{3+t^{2}}}\qquad {\text{where }}t=\tan {\frac {x}{2}}\\[6pt]&=\int _{-\infty }^{\infty }{\frac {2\,dt}{3+t^{2}}}\\[6pt]&={\frac {2}{\sqrt {3}}}\int _{-\infty }^{\infty }{\frac {du}{1+u^{2}}}\qquad {\text{where }}t=u{\sqrt {3}}\\[6pt]&={\frac {2\pi }{\sqrt {3}}}.\end{aligned}}$ Note: The substitution $t=\tan {\frac {x}{2}}$ maps $x\in (0,\pi )$ to $t\in (0,\infty )$ and $x\in (\pi ,2\pi )$ to $t\in (-\infty ,0)$ . The point $x=\pi$ corresponds to a vertical asymptote in $t {\displaystyle t}$ , so the integral is evaluated as a limit around this point.

Alternatively, we can compute the indefinite integral first:

$\int {\frac {dx}{2+\cos x}}={\frac {2}{\sqrt {3}}}\arctan \left({\frac {\tan {\frac {x}{2}}}{\sqrt {3}}}\right)+C$ Using symmetry:

${\begin{aligned}\int _{0}^{2\pi }{\frac {dx}{2+\cos x}}&=2\int _{0}^{\pi }{\frac {dx}{2+\cos x}}\\[6pt]&=2\lim _{b\to \pi ^{-}}\left[{\frac {2}{\sqrt {3}}}\arctan \left({\frac {\tan {\frac {x}{2}}}{\sqrt {3}}}\right)\right]_{0}^{b}\\[6pt]&={\frac {4}{\sqrt {3}}}\left({\frac {\pi }{2}}-0\right)={\frac {2\pi }{\sqrt {3}}}.\end{aligned}}$ Thus, the value of the definite integral is:

$\int _{0}^{2\pi }{\frac {dx}{2+\cos x}}={\frac {2\pi }{\sqrt {3}}}$

Third example: both sine and cosine

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${\begin{aligned}\int {\frac {dx}{a\cos x+b\sin x+c}}&=\int {\frac {2\,dt}{a(1-t^{2})+2bt+c(t^{2}+1)}}\\[6pt]&=\int {\frac {2\,dt}{(c-a)t^{2}+2bt+a+c}}\\[6pt]&={\frac {2}{\sqrt {c^{2}-(a^{2}+b^{2})}}}\arctan \left({\frac {(c-a)\tan {\tfrac {x}{2}}+b}{\sqrt {c^{2}-(a^{2}+b^{2})}}}\right)+C\end{aligned}}$ if ${\textstyle c^{2}-(a^{2}+b^{2})>0.}$

Geometry

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The tangent half-angle substitution parametrizes theunit circle centered at (0, 0). Instead of +∞ and −∞, we have only one ∞, at both ends of the real line. That is often appropriate when dealing with rational functions and with trigonometric functions. (This is theone-point compactification of the line.)

Asx varies, the point (cos x, sin x) winds repeatedly around theunit circle centered at (0, 0). The point

$\left({\frac {1-t^{2}}{1+t^{2}}},{\frac {2t}{1+t^{2}}}\right)$

goes only once around the circle ast goes from −∞ to +∞, and never reaches the point (−1, 0), which is approached as a limit ast approaches ±∞. Ast goes from −∞ to −1, the point determined byt goes through the part of the circle in the third quadrant, from (−1, 0) to (0, −1). Ast goes from −1 to 0, the point follows the part of the circle in the fourth quadrant from (0, −1) to (1, 0). Ast goes from 0 to 1, the point follows the part of the circle in the first quadrant from (1, 0) to (0, 1). Finally, ast goes from 1 to +∞, the point follows the part of the circle in the second quadrant from (0, 1) to (−1, 0).

Here is another geometric point of view. Draw the unit circle, and letP be the point(−1, 0). A line throughP (except the vertical line) is determined by its slope. Furthermore, each of the lines (except the vertical line) intersects the unit circle in exactly two points, one of which isP. This determines a function from points on the unit circle to slopes. 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.

Hyperbolic functions

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As with other properties shared between the trigonometric functions and the hyperbolic functions, it is possible to usehyperbolic identities to construct a similar form of the substitution, ${\textstyle t=\tanh {\tfrac {x}{2}}}$ :

$\sinh x={\frac {2t}{1-t^{2}}},\quad \cosh x={\frac {1+t^{2}}{1-t^{2}}},\quad {\text{and}}\quad dx={\frac {2}{1-t^{2}}}\,dt.$

Similar expressions can be written fortanhx,cothx,sechx, andcschx. Geometrically, this change of variables is a one-dimensional stereographic projection of thehyperbolic line onto the real interval, analogous to thePoincaré disk model of the hyperbolic plane.

Alternatives

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There are other approaches to integrating trigonometric functions. For example, it can be helpful to rewrite trigonometric functions in terms ofe^ix ande^−ix usingEuler's formula.

Notes and references

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^Other trigonometric functions can be written in terms of sine and cosine.
^Gunter, Edmund (1673) [1624].The Works of Edmund Gunter. Francis Eglesfield.p. 73
^Euler, Leonhard (1768)."§1.1.5.261 Problema 29"(PDF).Institutiones calculi integralis [Foundations of Integral Calculus] (in Latin). Vol. I. Impensis Academiae Imperialis Scientiarum. pp. 148–150.E 342,Translation by Ian Bruce.
Also seeLobatto, Rehuel (1832)."19. Note sur l'intégration de la fonction∂z / (a +b cosz)".Crelle's Journal (in French).9:259–260.
^Legendre, Adrien-Marie (1817).Exercices de calcul intégral [Exercises in integral calculus] (in French). Vol. 2. Courcier.p. 245–246.
^
For example, in chronological order,
- Hermite, Charles (1873).Cours d'analyse de l'école polytechnique (in French). Vol. 1. Gauthier-Villars. p. 320.
- Johnson, William Woolsey (1883).An Elementary Treatise on the Integral Calculus. MacMillan. p. 52.
- Picard, Émile (1901) [1891].Traité d'analyse (in French). Vol. 1 (2e ed.). Gauthier-Villars. p. 77.
- Goursat, Édouard (1904) [1902].A Course in Mathematical Analysis. Vol. 1. Translated by Hedrick, Earle Raymond. Ginn. pp. 236ff.
- Wilson, Edwin Bidwell (1911).Advanced Calculus. Ginn. p. 21.
- Edwards, Joseph (1921).A Treatise on the Integral Calculus. Vol. 1. MacMillan. pp. 187–188.
- Courant, Richard (1937) [1930].Differential and Integral Calculus. Vol. 1. Translated by McShane, E. J. (2nd ed.). Blackie & Son. pp. 234–238.
- Peterson, Thurman S. (1950).Elements of Calculus. Harper & Brothers. pp. 201–202.
- Apostol, Tom M. (1967) [1961].Calculus. Vol. 1 (2nd ed.). Xerox. pp. 264–265.
- Larson, Ron; Hostetler, Robert P.; Edwards, Bruce H. (1998) [1978].Calculus of a Single Variable (6th ed.). Houghton Mifflin. p. 520.ISBN 9780395885789.
- Rogawski, Jon (2011) [2008].Calculus: Early Transcendentals (2nd ed.). Macmillan. p. 435.ISBN 9781429231848.
^Piskunov, Nikolai (1969).Differential and Integral Calculus. Mir. p. 379.
Zaitsev, V. V.; Ryzhkov, V. V.; Skanavi, M. I. (1978).Elementary Mathematics: A Review Course. Ėlementarnai͡a matematika.English. Mir. p. 388.
^
In 1966William Eberlein attributed this substitution toKarl Weierstrass (1815–1897):
Eberlein, William Frederick (1966). "The Circular Function(s)".Mathematics Magazine.39 (4):197–201.doi:10.1080/0025570X.1966.11975715.JSTOR 2688079.(Equations (3)[ $x=\cos \theta$ ], (4)[ $y=\sin \theta$ ], (5)[ $t=\tan {\tfrac {\theta }{2}}$ ] are, of course, the familiar half-angle substitutions introduced by Weierstrass to integrate rational functions of sine, cosine.)
Two decades later,James Stewart mentioned Weierstrass when discussing the substitution in his popular calculus textbook, first published in 1987:
Stewart, James (1987)."§7.5 Rationalizing substitutions".Calculus. Brooks/Cole. p. 431.ISBN 9780534066901.The German mathematician Karl Weierstrass (1815–1897) noticed that the substitutiont = tan(x/2) will convert any rational function ofsinx andcosx into an ordinary rational function.
Later authors, citing Stewart, have sometimes referred to this as theWeierstrass substitution, for instance:
- Jeffrey, David J.; Rich, Albert D. (1994)."The evaluation of trigonometric integrals avoiding spurious discontinuities".Transactions on Mathematical Software.20 (1):124–135.doi:10.1145/174603.174409.S2CID 13891212.
- Merlet, Jean-Pierre (2004)."A Note on the History of Trigonometric Functions"(PDF). In Ceccarelli, Marco (ed.).International Symposium on History of Machines and Mechanisms. Kluwer. pp. 195–200.doi:10.1007/1-4020-2204-2_16.ISBN 978-1-4020-2203-6.
- Weisstein, Eric W. (2011)."Weierstrass Substitution".MathWorld. Retrieved2020-04-01.
Neither Eberlein nor Stewart provided any evidence for the attribution to Weierstrass. A related substitution appears in Weierstrass’sMathematical Works, from an 1875 lecture wherein Weierstrass creditsCarl Gauss (1818) with the idea of solving an integral of the form ${\textstyle \int d\psi \,H(\sin \psi ,\cos \psi ){\big /}{\sqrt {G(\sin \psi ,\cos \psi )}}}$ by the substitution ${\textstyle t=-\cot(\psi /2).}$
Weierstrass, Karl (1915) [1875]."8. Bestimmung des Integrals ...".Mathematische Werke von Karl Weierstrass (in German). Vol. 6. Mayer & Müller. pp. 89–99.
^Spivak, Michael (1967)."Ch. 9, problems 9–10".Calculus. Benjamin. pp. 325–326.

External links

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Weierstrass substitution formulas atPlanetMath

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Integration techniques	Substitution Trigonometric Euler Weierstrass By parts Partial fractions Euler's formula Inverse functions Changing order Reduction formulas Parametric derivatives Differentiation under the integral sign Laplace transform Contour integration Laplace's method Numerical integration Simpson's rule Trapezoidal rule Risch algorithm
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