Tangent half-angle substitution

Examples

Antiderivative of cosecant

$${\displaystyle \begin{array}{cc}\int \csc xdx& =\int \frac{dx}{\sin x}\\ & =\int \left(\frac{1+t^2}{2t}\right)\left(\frac{2}{1+t^2}\right)dt& & t=\tan \frac{x}{2}\\ & =\int \frac{dt}{t}\\ & =\ln |t|+C\\ & =\ln \left|\tan \frac{x}{2}\right|+C.\end{array}}$$

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

These two answers are the same because $\csc x-\cot x=\tan \frac{x}{2}:$

$${\displaystyle \begin{array}{cc}\csc x-\cot x& =\frac{1}{\sin x}-\frac{\cos x}{\sin x}\\ & =\frac{1+t^2}{2t}-\frac{1-t^2}{1+t^2}\frac{1+t^2}{2t}t=\tan \frac{x}{2}\\ & =\frac{2t^2}{2t}=t\\ & =\tan \frac{x}{2}\end{array}}$$

The secant integral may be evaluated in a similar manner.

A definite integral

We wish to evaluate the integral:

${\displaystyle {\int }_0^{2\pi }\frac{dx}{2+\cos x}}$

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

${\displaystyle \begin{array}{cc}{\int }_0^{2\pi }\frac{dx}{2+\cos x}& ={\int }_0^{\pi }\frac{dx}{2+\cos x}+{\int }_{\pi }^{2\pi }\frac{dx}{2+\cos x}\\ & ={\int }_0^{\mathrm{\infty }}\frac{2dt}{3+t^2}+{\int }_{-\mathrm{\infty }}^0\frac{2dt}{3+t^2}\text{where }t=\tan \frac{x}{2}\\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{2dt}{3+t^2}\\ & =\frac{2}{\sqrt{3}}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{du}{1+u^2}\text{where }t=u\sqrt{3}\\ & =\frac{2\pi }{\sqrt{3}}.\end{array}}$ Note: The substitution ${\displaystyle t=\tan \frac{x}{2}}$ maps ${\displaystyle x\in (0,\pi )}$ to ${\displaystyle t\in (0,\mathrm{\infty })}$ and ${\displaystyle x\in (\pi ,2\pi )}$ to ${\displaystyle t\in (-\mathrm{\infty },0)}$. The point ${\displaystyle x=\pi }$ corresponds to a vertical asymptote in ${\displaystyle t}$, so the integral is evaluated as a limit around this point.

Alternatively, we can compute the indefinite integral first:

${\displaystyle \frac{2}{\sqrt{3}}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{du}{1+u^2}}$

this integral evaluates to arctan(u) after solving for u by using u=tan(z) as a substitution then solving for t gives us

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

Using symmetry:

${\displaystyle \begin{array}{cc}{\int }_0^{2\pi }\frac{dx}{2+\cos x}& =2{\int }_0^{\pi }\frac{dx}{2+\cos x}\\ & =2\underset{b\to {\pi }^{-}}{lim}{\left[\frac{2}{\sqrt{3}}\arctan \left(\frac{\tan \frac{x}{2}}{\sqrt{3}}\right)\right]}_0^b\\ & =\frac{4}{\sqrt{3}}\left(\frac{\pi }{2}-0\right)=\frac{2\pi }{\sqrt{3}}.\end{array}}$ Thus, the value of the definite integral is:

${\displaystyle {\int }_0^{2\pi }\frac{dx}{2+\cos x}=\frac{2\pi }{\sqrt{3}}}$

Third example: both sine and cosine

$${\displaystyle \begin{array}{cc}\int \frac{dx}{a\cos x+b\sin x+c}& =\int \frac{2dt}{a(1-t^2)+2bt+c(t^2+1)}\\ & =\int \frac{2dt}{(c-a)t^2+2bt+a+c}\\ & =\frac{2}{\sqrt{c^2-(a^2+b^2)}}\arctan \left(\frac{(c-a)\tan \frac{x}{2}+b}{\sqrt{c^2-(a^2+b^2)}}\right)+C\end{array}}$$ if $c^2-(a^2+b^2)>0.$

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, $t=\tanh \frac{x}{2}$:

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

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.