Tangent half-angle substitution

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 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 in spherical trigonometry and was sometimes known in the 17th century as the half tangent or semi-tangent. Leonhard Euler used it to evaluate the integral ${\textstyle \int dx/(a+b\cos x)}$ in his 1768 integral calculus textbook, and Adrien-Marie Legendre described the general method in 1817.

The substitution is described in most integral calculus textbooks since the late 19th century, usually without any special name. It is known in Russia as the universal trigonometric substitution, and also known by variant names such as half-tangent substitution or half-angle substitution. It is sometimes misattributed as the Weierstrass substitution. Michael Spivak called it the "world's sneakiest substitution".

The substitution

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 as rational functions of $t,$ and $dx$ can be expressed as the product of $dt$ and a rational function of $t,$ 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 for tan x, cot x, sec x, and csc x.

Derivation

Using the double-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 the Pythagorean 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

Antiderivative of cosecant

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

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

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

The secant integral may be evaluated in a similar manner.

A definite integral

$${\begin{aligned}\int _{0}^{2\pi }{\frac {dx}{2+\cos x}}&=\int _{0}^{\pi }{\frac {dx}{2+\cos x}}+\int _{\pi }^{2\pi }{\frac {dx}{2+\cos x}}\\[6pt]&=\int _{0}^{\infty }{\frac {2\,dt}{3+t^{2}}}+\int _{-\infty }^{0}{\frac {2\,dt}{3+t^{2}}}&t&=\tan {\tfrac {x}{2}}\\[6pt]&=\int _{-\infty }^{\infty }{\frac {2\,dt}{3+t^{2}}}\\[6pt]&={\frac {2}{\sqrt {3}}}\int _{-\infty }^{\infty }{\frac {du}{1+u^{2}}}&t&=u{\sqrt {3}}\\[6pt]&={\frac {2\pi }{\sqrt {3}}}.\end{aligned}}$$

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. $${\begin{aligned}\int {\frac {dx}{2+\cos x}}&=\int {\frac {1}{2+{\frac {1-t^{2}}{1+t^{2}}}}}{\frac {2\,dt}{t^{2}+1}}&&t=\tan {\tfrac {x}{2}}\\[6pt]&=\int {\frac {2\,dt}{2(t^{2}+1)+(1-t^{2})}}=\int {\frac {2\,dt}{t^{2}+3}}\\[6pt]&={\frac {2}{3}}\int {\frac {dt}{{\bigl (}t{\big /}{\sqrt {3}}{\bigr )}^{2}+1}}&&u=t{\big /}{\sqrt {3}}\\[6pt]&={\frac {2}{\sqrt {3}}}\int {\frac {du}{u^{2}+1}}&&\tan \theta =u\\[6pt]&={\frac {2}{\sqrt {3}}}\int \cos ^{2}\theta \sec ^{2}\theta \,d\theta ={\frac {2}{\sqrt {3}}}\int d\theta \\[6pt]&={\frac {2}{\sqrt {3}}}\theta +C={\frac {2}{\sqrt {3}}}\arctan \left({\frac {t}{\sqrt {3}}}\right)+C\\[6pt]&={\frac {2}{\sqrt {3}}}\arctan \left({\frac {\tan {\tfrac {x}{2}}}{\sqrt {3}}}\right)+C.\end{aligned}}$$ By symmetry, $${\begin{aligned}\int _{0}^{2\pi }{\frac {dx}{2+\cos x}}&=2\int _{0}^{\pi }{\frac {dx}{2+\cos x}}=\lim _{b\rightarrow \pi }{\frac {4}{\sqrt {3}}}\arctan \left({\frac {\tan {\tfrac {x}{2}}}{\sqrt {3}}}\right){\Biggl |}_{0}^{b}\\[6pt]&={\frac {4}{\sqrt {3}}}{\Biggl [}\lim _{b\rightarrow \pi }\arctan \left({\frac {\tan {\tfrac {b}{2}}}{\sqrt {3}}}\right)-\arctan(0){\Biggl ]}={\frac {4}{\sqrt {3}}}\left({\frac {\pi }{2}}-0\right)={\frac {2\pi }{\sqrt {3}}},\end{aligned}}$$ which is the same as the previous answer.

Third example: both sine and cosine

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

Geometry

The tangent half-angle substitution parametrizes the unit 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 the one-point compactification of the line.)

As x varies, the point (cos x, sin x) winds repeatedly around the unit 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 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, ${\textstyle t=\tanh {\tfrac {x}{2}}}$:

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

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 e^ix and e^−ix using Euler's formula.

See also

• Rational curve
• Stereographic projection
• Tangent half-angle formula
• Trigonometric substitution
• Euler substitution

Further reading

• 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.

External links

• Weierstrass substitution formulas at PlanetMath