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 ${\textstyle x}$ into an ordinary rational function of ${\textstyle t}$ by setting ${\textstyle t=\tan {\tfrac {x}{2))}$. This is the one-dimensional stereographic projection of the unit circle parametrized by angle measure onto the real line. The general[1] transformation formula is:

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

The tangent of half an angle is important in spherical 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 his 1768 integral calculus textbook,[3] and Adrien-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 the universal trigonometric substitution,[6] and also known by variant names such as half-tangent substitution or half-angle substitution. It is sometimes misattributed as the Weierstrass substitution.[7] Michael Spivak called it the "world's sneakiest substitution".[8]

The substitution

Introducing a new variable ${\textstyle t=\tan {\tfrac {x}{2)),}$ sines and cosines can be expressed as rational functions of ${\displaystyle t,}$ and ${\displaystyle dx}$ can be expressed as the product of ${\displaystyle dt}$ and a rational function of ${\displaystyle t,}$ as follows:

${\displaystyle \sin x={\frac {2t}{1+t^{2))},\quad \cos x={\frac {1-t^{2)){1+t^{2))},\quad {\text{and))\quad dx={\frac {2}{1+t^{2))}\,dt.}$

Similar expressions can be written for tan x, cot x, sec x, and csc x.

Derivation

Using the double-angle formulas ${\displaystyle \sin x=2\sin {\tfrac {x}{2))\cos {\tfrac {x}{2))}$ and ${\displaystyle \cos x=\cos ^{2}{\tfrac {x}{2))-\sin ^{2}{\tfrac {x}{2))}$ and introducing denominators equal to one by the Pythagorean identity ${\displaystyle 1=\cos ^{2}{\tfrac {x}{2))+\sin ^{2}{\tfrac {x}{2))}$ results in

{\displaystyle {\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

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

Examples

Antiderivative of cosecant

{\displaystyle {\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 ${\textstyle u=\csc x-\cot x,}$ ${\textstyle du=\left(-\csc x\cot x+\csc ^{2}x\right)\,dx}$.

{\displaystyle {\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 ${\textstyle \csc x-\cot x=\tan {\tfrac {x}{2))\colon }$

{\displaystyle {\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))}

The secant integral may be evaluated in a similar manner.

A definite integral

{\displaystyle {\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))}&t&=\tan {\tfrac {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))}&t&=u{\sqrt {3))\\[6pt]&={\frac {2\pi }{\sqrt {3))}.\end{aligned))}

In the first line, one cannot simply substitute ${\textstyle t=0}$ for both limits of integration. The singularity (in this case, a vertical asymptote) of ${\textstyle t=\tan {\tfrac {x}{2))}$ at ${\textstyle x=\pi }$ must be taken into account. Alternatively, first evaluate the indefinite integral, then apply the boundary values.

{\displaystyle {\begin{aligned}\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 {x}{2))\\[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 {x}{2))}{\sqrt {3))}\right)+C.\end{aligned))}
By symmetry,
{\displaystyle {\begin{aligned}\int _{0}^{2\pi }{\frac {dx}{2+\cos x))&=2\int _{0}^{\pi }{\frac {dx}{2+\cos x))=\lim _{b\rightarrow \pi }{\frac {4}{\sqrt {3))}\arctan \left({\frac {\tan {\tfrac {x}{2))}{\sqrt {3))}\right){\Biggl |}_{0}^{b}\\[6pt]&={\frac {4}{\sqrt {3))}{\Biggl [}\lim _{b\rightarrow \pi }\arctan \left({\frac {\tan {\tfrac {b}{2))}{\sqrt {3))}\right)-\arctan(0){\Biggl ]}={\frac {4}{\sqrt {3))}\left({\frac {\pi }{2))-0\right)={\frac {2\pi }{\sqrt {3))},\end{aligned))}
which is the same as the previous answer.

Third example: both sine and cosine

{\displaystyle {\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

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

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

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

Here is another geometric point of view. Draw the unit circle, and let P be the point (−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 hyperbolic identities to construct a similar form of the substitution, ${\textstyle t=\tanh {\tfrac {x}{2))}$:

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

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

Alternatives

There are other approaches to integrating trigonometric functions. For example, it can be helpful to rewrite trigonometric functions in terms of eix and eix using Euler's formula.

• Courant, Richard (1937) [1934]. "1.4.6. Integration of Some Other Classes of Functions §1–3". Differential and Integral Calculus. Vol. 1. Blackie & Son. pp. 234–237.
• Edwards, Joseph (1921). "§1.6.193". A Treatise on the Integral Calculus. Vol. 1. Macmillan. pp. 187–188.
• Hardy, Godfrey Harold (1905). "VI. Transcendental functions". The integration of functions of a single variable. Cambridge. pp. 42–51. Second edition 1916, pp. 52–62
• Hermite, Charles (1873). "Intégration des fonctions transcendentes" [Integration of transcendental functions]. Cours d'analyse de l'école polytechnique (in French). Vol. 1. Gauthier-Villars. pp. 320–380.
• Stewart, Seán M. (2017). "14. Tangent Half-Angle Substitution". How to Integrate It. Cambridge. pp. 178–187. doi:10.1017/9781108291507.015. ISBN 978-1-108-41881-2.

Notes and references

1. ^ Other trigonometric functions can be written in terms of sine and cosine.
2. ^ Gunter, Edmund (1673) [1624]. The Works of Edmund Gunter. Francis Eglesfield. p. 73
3. ^ 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. E342, Translation by Ian Bruce.
Also see Lobatto, Rehuel (1832). "19. Note sur l'intégration de la fonction z/(a + b cos z)". Crelle's Journal (in French). 9: 259–260.
4. ^ Legendre, Adrien-Marie (1817). Exercices de calcul intégral [Exercises in integral calculus] (in French). Vol. 2. Courcier. p. 245–246.
5. ^ For example, in chronological order,
6. ^ 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. Mir. p. 388.
7. ^ In 1966 William Eberlein attributed this substitution to Karl 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) [${\displaystyle x=\cos \theta }$], (4) [${\displaystyle y=\sin \theta }$], (5) [${\displaystyle t=\tan {\tfrac {\theta }{2))}$] are, of course, the familiar half-angle substitutions introduced by Weierstrass to integrate rational functions of sine, cosine.)
Two decades later, James Stewart mentioned Weirstrass 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 substitution t = tan(x/2) will convert any rational function of sin x and cos x into an ordinary rational function.

Later authors, citing Stewart, have sometimes referred to this as the Weierstrass substitution, for instance:

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 Carl 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.

8. ^ Spivak, Michael (1967). "Ch. 9, problems 9–10". Calculus. Benjamin. pp. 325–326.