View source for Tangent half-angle substitution

{{short description|Change of variable for integrals involving trigonometric functions}}
{{calculus|expanded=integral}}

In integral calculus, the '''tangent half-angle substitution''' is a [[Integration by substitution|change of variables]] used for evaluating [[Antiderivative|integrals]], which converts a [[Rational function|rational function]] of [[Trigonometric functions|trigonometric functions]] of <math display=inline>x</math> into an ordinary rational function of <math display=inline>t</math> by setting <math display=inline>t = \tan \tfrac x2</math>. This is the one-dimensional [[Stereographic projection|stereographic projection]] of the [[Unit circle|unit circle]] parametrized by angle measure onto the [[Real line|real line]]. The general<ref>Other trigonometric functions can be written in terms of sine and cosine.</ref> transformation formula is:

<math display=block>\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}.</math>

The tangent of half an angle is important in [[Spherical trigonometry|spherical trigonometry]] and was sometimes known in the 17th century as the half tangent or semi-tangent.<ref>{{cite book |last=Gunter |first=Edmund |year=1673 |orig-year=1624 |title=The Works of Edmund Gunter |publisher=Francis Eglesfield |url=https://archive.org/details/worksofedmundgun00gunt/}} [https://archive.org/details/worksofedmundgun00gunt/page/n104/ p. 73]</ref> [[Biography:Leonhard Euler|Leonhard Euler]] used it to evaluate the integral <math display=inline>\int dx / (a + b\cos x)</math> in his [[Institutiones calculi integralis|1768 integral calculus textbook]],<ref>{{Cite book |last=Euler |first=Leonhard |date=1768 |title=[[Institutiones calculi integralis]] |volume=I |lang=la |trans-title=Foundations of Integral Calculus |chapter=§1.1.5.261 Problema 29 |publisher=Impensis Academiae Imperialis Scientiarum |pages=148–150 |chapter-url=http://eulerarchive.maa.org/docs/originals/E342sec1ch5.pdf }} [[Biography:Leonhard Euler#Selected bibliography|E]][https://scholarlycommons.pacific.edu/euler-works/342/ 342], [http://www.17centurymaths.com/contents/euler/intcalvol1/part1ch5.pdf Translation by Ian Bruce].
<br/>Also see {{cite journal |last=Lobatto |first=Rehuel |year=1832 |title=19. Note sur l'intégration de la fonction {{math|∂''z''{{thinsp}}/{{thinsp}}(''a'' + ''b'' cos ''z'')}} |lang=fr |journal=Crelle's Journal |volume=9 |pages=259–260 |url=https://archive.org/details/journalfurdierei9101unse/page/n268 }}</ref> and [[Biography:Adrien-Marie Legendre|Adrien-Marie Legendre]] described the general method in 1817.<ref>{{cite book |last=Legendre |first=Adrien-Marie |title=Exercices de calcul intégral |volume=2 |trans-title=Exercises in integral calculus |lang=fr |publisher=Courcier |year=1817 |url=https://archive.org/details/exercicescalculi02legerich }} [https://archive.org/details/exercicescalculi02legerich/page/n267/ p. 245–246].</ref>

The substitution is described in most integral calculus textbooks since the late 19th century, usually without any special name.{{r|unnamed}} It is known in Russia as the '''universal trigonometric substitution''',<ref>{{cite book |last=Piskunov |first=Nikolai |title=Differential and Integral Calculus |publisher=Mir |year=1969 |page=379 |url=https://archive.org/details/n.-piskunov-differential-and-integral-calculus-mir-1969/page/379/ }} {{pb}} {{cite book |last1=Zaitsev |first1=V. V. |last2=Ryzhkov |first2=V. V. |last3=Skanavi |first3=M. I. |year=1978 |title=Elementary Mathematics: A Review Course |publisher=Mir |page=388 |url=https://catalog.hathitrust.org/Record/000248796 |url-access=subscription }} </ref> and also known by variant names such as ''half-tangent substitution'' or ''half-angle substitution''. It is sometimes misattributed as the '''Weierstrass substitution'''.{{r|weierstrass}} [[Biography:Michael Spivak|Michael Spivak]] called it the "world's sneakiest substitution".<ref>{{cite book |last=Spivak |first=Michael |year=1967 |title=Calculus |chapter=Ch. 9, problems 9–10 |chapter-url=https://archive.org/details/CalculusSpivak/page/n337/ |publisher=Benjamin |pages=325–326}}</ref>

== The substitution ==
[[File:Weierstrass substitution.svg|thumb|upright=1.25|The tangent half-angle substitution relates an angle to the slope of a line.]]

Introducing a new variable <math display=inline>t=\tan\tfrac x2,</math> sines and cosines can be expressed as [[Rational function|rational function]]s of <math>t,</math> and <math>dx</math> can be expressed as the product of <math>dt</math> and a rational function of <math>t,</math> as follows:
<math display=block>
\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.
</math>

Similar expressions can be written for {{math|tan ''x''}}, {{math|cot ''x''}}, {{math|sec ''x''}}, and {{math|csc ''x''}}.

=== Derivation ===
Using the double-angle formulas <math>\sin x = 2 \sin \tfrac x2 \cos \tfrac x2</math> and <math>\cos x = \cos^2 \tfrac x2 - \sin^2 \tfrac x2</math> and introducing denominators equal to one by the Pythagorean identity <math>1 = \cos^2 \tfrac x2 + \sin^2 \tfrac x2</math> results in

<math display=block>\begin{align}
\sin x &= \frac {2\sin \tfrac x2\, \cos \tfrac x2}{\cos^2\tfrac x2 + \sin^2 \tfrac x2} = \frac{2\tan \tfrac x2}{1+\tan^2 \tfrac x2} = \frac{2t}{1 + t^2}, \\[18mu]
\cos x &= \frac {\cos^2 \tfrac x2 - \sin^2 \tfrac x2}{\cos^2 \tfrac x2 + \sin^2 \tfrac x2} = \frac{1-\tan^2 \tfrac x2}{1 + \tan^2 \tfrac x2} = \frac{1 - t^2}{1 + t^2}.\end{align}</math>

Finally, since <math display=inline>t = \tan \tfrac x2 </math>, [[Differentiation rules|differentiation rules]] imply

<math display=block>dt = \tfrac12\left(1+\tan^2 \tfrac x2\right) dx = \frac{1+t^2}2 \, dx,</math>
and thus 
<math display=block>dx=\frac{2}{1 + t^2} \, dt.</math>

== Examples ==
=== Antiderivative of cosecant ===
<math display=block>\begin{align}
\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 x2 \\[6pt]
&=\int\frac{dt}{t} \\[6pt]
&=\ln |t |+ C \\[6pt]
&=\ln \left|\tan\tfrac x2 \right| + C.
\end{align}</math>

We can confirm the above result using a standard method of evaluating the cosecant integral by multiplying the numerator and denominator by <math display=inline>\csc x - \cot x</math> and performing the substitution <math display=inline>u = \csc x - \cot x,</math> <math display=inline>du = \left(-\csc x \cot x + \csc^2 x\right)\,dx</math>.
<math display=block>
\begin{align}
\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{align}
</math>

These two answers are the same because <math display=inline>\csc x - \cot x = \tan \tfrac x2\colon</math>

<math display=block>\begin{align}
\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 x2 \\[6pt]
&= \frac{2t^2}{2t} = t \\[6pt]
&= \tan \tfrac x2
\end{align}</math>

The [[Integral of the secant function|secant integral]] may be evaluated in a similar manner.

=== A definite integral ===
<math display=block>\begin{align}
\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} & t &= \tan\tfrac x2 \\[6pt]
&=\int_{-\infty}^\infty \frac{2\,dt}{3+t^2} \\[6pt]
&=\frac{2}{\sqrt 3}\int_{-\infty}^\infty \frac{du}{1+u^2} & t &= u\sqrt 3 \\[6pt]
&=\frac{2\pi}{\sqrt 3}.
\end{align}</math>

In the first line, one cannot simply substitute <math display=inline>t=0</math> for both [[Limits of integration|limits of integration]]. The singularity (in this case, a [[Asymptote#Vertical asymptotes|vertical asymptote]]) of <math display=inline>t=\tan\tfrac x2</math> at <math display=inline>x=\pi</math> must be taken into account. Alternatively, first evaluate the indefinite integral, then apply the boundary values.
<math display=block>\begin{align}
\int \frac{dx}{2 + \cos x} &= \int \frac{1}{2 + \frac{1-t^2}{1+t^2}} \frac{2\,dt}{t^2+1} && t = \tan\tfrac x2 \\[6pt]
&= \int \frac{2\, dt}{2(t^2+1)+(1-t^2)} = \int \frac{2\,dt}{t^2+3}\\[6pt]
&= \frac{2}{3} \int \frac{dt}{\bigl(t \big/ \sqrt 3\bigr)^2 + 1} && u = t \big/ \sqrt 3\\[6pt]
&= \frac{2}{\sqrt 3} \int \frac{du}{u^2 + 1} && \tan \theta = u \\[6pt]
&= \frac{2}{\sqrt 3} \int \cos^2 \theta \sec^2 \theta \,d\theta = \frac{2}{\sqrt 3} \int d\theta\\[6pt]
&= \frac{2}{\sqrt 3} \theta + C = \frac{2}{\sqrt 3} \arctan \left( \frac{t}{\sqrt 3}\right) + C\\[6pt]
&= \frac{2}{\sqrt 3} \arctan \left( \frac{\tan\tfrac x2}{\sqrt3}\right) + C.
\end{align}</math>
By symmetry,
<math display=block>\begin{align}
\int_{0}^{2\pi} \frac{dx}{2 + \cos x} &= 2 \int_{0}^{\pi} \frac{dx}{2 + \cos x} = \lim_{b \rightarrow \pi} \frac{4}{\sqrt3} \arctan \left( \frac{\tan\tfrac x2}{\sqrt3}\right) \Biggl|_{0}^{b}\\[6pt]
&= \frac{4}{\sqrt3} \Biggl[ \lim_{b \rightarrow \pi} \arctan \left(\frac{\tan\tfrac b2}{\sqrt3}\right) -  \arctan (0) \Biggl] = \frac{4}{\sqrt 3} \left( \frac{\pi}{2} - 0\right) = \frac{2\pi}{\sqrt 3},
\end{align} </math>
which is the same as the previous answer.

===Third example: both sine and cosine===
<math display=block>\begin{align} \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 x2 + b}{\sqrt{c^2-(a^2+b^2)}}\right) + C
\end{align} </math>
if <math display=inline> c^2-(a^2+b^2)>0.</math>

== Geometry ==
[[File:Weierstrass.substitution.svg|thumb|upright=1.25|The tangent half-angle substitution parametrizes the [[Unit circle|unit circle]] centered at&nbsp;(0,&nbsp;0).  Instead of +&infin; and &minus;&infin;, we have only one &infin;, at both ends of the real line.  That is often appropriate when dealing with rational functions and with trigonometric functions.  (This is the one-point compactification of the line.)|alt=]]

As ''x'' varies, the point (cos&nbsp;''x'',&nbsp;sin&nbsp;''x'') winds repeatedly around the [[Unit circle|unit circle]] centered at&nbsp;(0,&nbsp;0).  The point

<math display=block>\left(\frac{1-t^2}{1+t^2}, \frac{2t}{1+t^2}\right)</math>

goes only once around the circle as ''t'' goes from &minus;&infin; to&nbsp;+&infin;, and never reaches the point&nbsp;(&minus;1,&nbsp;0), which is approached as a limit as ''t'' approaches&nbsp;±&infin;.  As ''t'' goes from &minus;&infin; to &minus;1, the point determined by ''t'' goes through the part of the circle in the third quadrant, from (&minus;1,&nbsp;0) to&nbsp;(0,&nbsp;&minus;1).  As ''t'' goes from &minus;1 to&nbsp;0, the point follows the part of the circle in the fourth quadrant from (0,&nbsp;&minus;1) to&nbsp;(1,&nbsp;0).  As ''t'' goes from 0 to 1, the point follows the part of the circle in the first quadrant from (1,&nbsp;0) to&nbsp;(0,&nbsp;1).  Finally, as ''t'' goes from 1 to&nbsp;+&infin;, the point follows the part of the circle in the second quadrant from (0,&nbsp;1) to&nbsp;(&minus;1,&nbsp;0).

Here is another geometric point of view.  Draw the unit circle, and let ''P'' be the point {{nowrap|(&minus;1, 0)}}.  A line through ''P'' (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 is ''P''.  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==

As with other properties shared between the trigonometric functions and the hyperbolic functions, it is possible to use [[Tangent half-angle formula#Hyperbolic identities|hyperbolic identities]] to construct a similar form of the substitution, <math display=inline>t = \tanh \tfrac x2</math>:

<math display=block>
\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.
</math>

Similar expressions can be written for {{math|tanh ''x''}}, {{math|coth ''x''}}, {{math|sech ''x''}}, and {{math|csch ''x''}}.  Geometrically, this change of variables is a one-dimensional stereographic projection of the [[Hyperbolic space|hyperbolic]] line onto the real interval, analogous to the [[Poincaré disk model]] of the hyperbolic plane.

==See also==
*Rational curve
*[[Stereographic projection]]
*[[Tangent half-angle formula]]
*[[Trigonometric substitution]]
*[[Euler substitution]]

== Further reading ==

* {{cite book |last=Courant |first=Richard |year=1937 |orig-year=1934 |title=Differential and Integral Calculus |volume=1 |publisher=Blackie & Son |chapter=1.4.6. Integration of Some Other Classes of Functions §1–3 |pages=234–237 |chapter-url=https://archive.org/details/differentialinte0001cour/page/234 |chapter-url-access=registration }}
* {{Cite book |last=Edwards |first=Joseph |year=1921 |title=A Treatise on the Integral Calculus |volume=1 |publisher=Macmillan |chapter=§1.6.193 |chapter-url=https://archive.org/details/treatiseonintegr01edwauoft/page/187 |pages=187–188}}
* {{cite book |last=Hardy |first=Godfrey Harold |year=1905 |title=The integration of functions of a single variable |chapter=VI. Transcendental functions |publisher=Cambridge |chapter-url=https://archive.org/details/integrationoffun00hardrich/page/42 |pages=42–51}} Second edition 1916, [https://archive.org/details/integrationoffun00hard/page/52/ pp. 52–62]
* {{cite book |last=Hermite |first=Charles |year=1873 |title=Cours d'analyse de l'école polytechnique |volume=1 |chapter=Intégration des fonctions transcendentes |trans-chapter=Integration of transcendental functions |lang=fr |publisher=Gauthier-Villars |chapter-url=https://archive.org/details/coursdanalysedel01hermuoft/page/320 |pages=320–380 }}

== Notes and references ==
{{reflist|30em |refs=

<ref name=unnamed>For example, in chronological order, 
{{ulist
}}
}}
}}
}}
}}
| {{cite book |last=Edwards |first=Joseph |year=1921 |title=A Treatise on the Integral Calculus |volume=1 |publisher=MacMillan |pages=187–188 |url=https://archive.org/details/treatiseonintegr01edwauoft/page/188
}}
}}
| {{cite book |last=Peterson |first=Thurman S.  |year=1950 |title=Elements of Calculus |publisher=Harper & Brothers |pages=201–202 |url=https://archive.org/details/elementsofcalcul00pete/page/201/ |url-access=limited
}}
}}
}}
| {{cite book |last=Rogawski |first=Jon |year=2011 |orig-year=2008 |title=Calculus: Early Transcendentals |edition=2nd |publisher=Macmillan |page=435 |isbn=9781429231848 |url=https://books.google.com/books?id=rn4paEb8izYC&pg=PA435 |url-access=limited
}}
}}
</ref>

<ref name=weierstrass>
In 1966 [[Biography:William Frederick Eberlein|William Eberlein]] attributed this substitution to [[Biography:Karl Weierstrass|Karl Weierstrass]] (1815–1897): {{pb}}
{{cite journal |last=Eberlein |first=William Frederick |year=1966 |title=The Circular Function(s) |journal=Mathematics Magazine |volume=39 |number=4 |pages=197–201 |doi=10.1080/0025570X.1966.11975715 |jstor=2688079 |quote=(Equations (3) {{nobr|[<math>x = \cos \theta</math>],}} (4) {{nobr|[<math>y = \sin \theta</math>],}} (5) {{nobr|[<math>t = \tan \tfrac\theta2</math>]}} are, of course, the familiar half-angle substitutions introduced by Weierstrass to integrate rational functions of sine, cosine.) }} {{pb}}
Two decades later, James Stewart mentioned Weirstrass when discussing the substitution in his popular calculus textbook, first published in 1987:{{pb}}
{{cite book |last=Stewart |first=James |title=Calculus |chapter=§7.5 Rationalizing substitutions |publisher=Brooks/Cole |year=1987 |page=431 |isbn=9780534066901 |chapter-url=https://archive.org/details/calculus00stew_3/page/431 |chapter-url-access=registration |quote=The German mathematician Karl Weierstrass (1815–1897) noticed that the substitution {{math|''t'' {{=}} tan(''x''/2)}} will convert any rational function of {{math|sin ''x''}} and {{math|cos ''x''}} into an ordinary rational function. }} {{pb}}
Later authors, citing Stewart, have sometimes referred to this as the ''Weierstrass substitution'', for instance:
{{ulist
| {{cite journal |last1=Jeffrey |first1=David J. |last2=Rich |first2=Albert D. |year=1994 |title=The evaluation of trigonometric integrals avoiding spurious discontinuities |journal=Transactions on Mathematical Software |volume=20 |issue=1 |pages=124–135 |doi=10.1145/174603.174409 |s2cid=13891212 |doi-access=free
}}
| {{cite book |last=Merlet |first=Jean-Pierre |year=2004 |chapter=A Note on the History of Trigonometric Functions |editor-last=Ceccarelli |editor-first=Marco |title=International Symposium on History of Machines and Mechanisms |publisher=Kluwer |pages=195–200 |doi=10.1007/1-4020-2204-2_16 |isbn=978-1-4020-2203-6 |chapter-url=https://ia800201.us.archive.org/15/items/springer_10.1007-1-4020-2204-2/10.1007-1-4020-2204-2.pdf#page=205
}}
}} }} {{pb}}
Neither Eberlein nor Stewart provided any evidence for the attribution to Weierstrass. A related substitution appears in Weierstrass’s ''Mathematical Works'', from an 1875 lecture wherein Weierstrass credits [[Biography:Carl Friedrich Gauss|Carl Gauss]] (1818) with the idea of solving an integral of the form <math display=inline>\int d\psi\, H(\sin \psi, \cos \psi) \big/ \sqrt{G(\sin \psi, \cos \psi)}</math> by the substitution <math display=inline>t = -\cot(\psi/2).</math> {{pb}}
{{cite book |last=Weierstrass |first=Karl |year=1915 |orig-date=1875 |publisher=Mayer & Müller |title=Mathematische Werke von Karl Weierstrass |lang=de |volume=6 |chapter=8. Bestimmung des Integrals ... |chapter-url=https://archive.org/details/mathematischewer06weieuoft/page/89 |pages=89–99 }}
</ref>

}} <!-- end reflist -->

== External links ==
* [https://planetmath.org/WeierstrassSubstitutionFormulas Weierstrass substitution formulas] at [[PlanetMath]]
{{Integrals}}
[[Category:Integral calculus]]

{{Sourceattribution|Tangent half-angle substitution}}