integration of rational function of sine and cosine

The integration task

${\displaystyle \int R(\sin x,\cos x)𝑑x},$ (1)

where the integrand is a rational function of $\sin x$ and $\cos x$, changes via the Weierstrass substitution

$\tan {\displaystyle \frac{x}{2}}=t$ (2)

to a form having an integrand that is a rational function of $t$.  Namely, since  $x=2\arctan t$,  we have

$dx=\mathrm{\hspace{0.33em}2}\cdot {\displaystyle \frac{1}{1+t^2}}dt,$ (3)

and we can substitute

$\sin x={\displaystyle \frac{2t}{1+t^2}},\cos x={\displaystyle \frac{1-t^2}{1+t^2}},$ (4)

getting

$$\int R(\sin x,\cos x)𝑑x=\mathrm{\hspace{0.33em}2}\int R(\frac{2t}{1+t^2},\frac{1-t^2}{1+t^2})\frac{dt}{1+t^2}.$$

Proof of the formulae (4):  Using the double angle formulas of sine and cosine and then dividing the numerators and the denominators by  ${\cos }^2\frac{x}{2}$  we obtain

$$\sin x=\frac{2\sin \frac{x}{2}\cos \frac{x}{2}}{{\sin }^2\frac{x}{2}+{\cos }^2\frac{x}{2}}=\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}=\frac{2t}{1+t^2},$$

$$\cos x=\frac{{\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2}}{{\sin }^2\frac{x}{2}+{\cos }^2\frac{x}{2}}=\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}=\frac{1-t^2}{1+t^2}.$$

Example.  The above formulae give from  ${\displaystyle \int \frac{dx}{\sin x}}$  the result

$$\int \frac{dx}{\sin x}=\int \frac{1+t^2}{2t}\cdot 2\cdot \frac{1}{1+t^2}𝑑t=\int \frac{dt}{t}=\ln |t|+C=\ln \left|\tan \frac{x}{2}\right|+C$$

(which can also be expressed in the form $-\ln |\csc x+\cot x|+C$; see the goniometric formulas).

Note 1.  The substitution (2) is sometimes called the ‘‘universal trigonometric substitution’’ (http://planetmath.org/UniversalTrigonometricSubstitution).  In practice, it often gives rational functions that are too complicated.  In many cases, it is more profitable to use other substitutions:

• In the case  $\int R(\sin x)\cos xdx$  the substitution  $\sin x=t$  is simpler.

• Similarly, in the case  $\int R(\cos x)\sin xdx$  the substitution  $\cos x=t$  is simpler.

• If the integrand depends only on $\tan x$, the substitution  $\tan x=t$  is simpler.

• If the integrand is of the form  $R({\sin }^{2}x,{\cos }^{2}x)$,  one can use the substitution  $\tan x=t$; then
  ${\cos }^{2}x={\displaystyle \frac{1}{1+{\tan }^{2}x}}={\displaystyle \frac{1}{1+t^2}}$,   ${\sin }^{2}x=1-{\cos }^{2}x={\displaystyle \frac{t^2}{1+t^2}}$,   $dx={\displaystyle \frac{dt}{1+t^2}}.$

Example.  The integration of  ${\displaystyle \int \frac{dx}{{\cos }^{4}x}𝑑x}$  is of the last case:

$$\int \frac{dx}{{\cos }^{4}x}𝑑x=\int \frac{1}{{({\cos }^{2}x)}^2}𝑑x=\int {(1+t^2)}^2\cdot \frac{dt}{1+t^2}=\int (1+t^2)𝑑t=\frac{t^3}{3}+t+C=\frac{1}{3}{\tan }^{3}x+\tan x+C.$$

Example.  The integral  $I={\displaystyle \int \frac{dx}{{\cos }^{3}x}𝑑x}={\displaystyle \int {\sec }^{3}xdx}$  is a peculiar case in which one does not have to use the substitutions mentioned above, as integration by parts is a simpler method for evaluating this integral. Thus,

$$u=\sec x\Rightarrow du=\sec x\tan xdx;dv={\sec }^{2}xdx\Rightarrow v=\tan x.$$

Therefore,

and consequently

$$\int \frac{dx}{{\cos }^{3}x}𝑑x=\frac{1}{2}\left(\sec x\tan x+\ln |\sec x+\tan x|\right)+C.$$

Note 2.  There is also the ‘‘universal hyperbolic substitution’’ for integrating rational functions of hyperbolic sine and cosine:

$$\tanh \frac{x}{2}=t,dx=\frac{2dt}{1-t^2},\sinh x=\frac{2t}{1-t^2},\cosh x=\frac{1+t^2}{1-t^2}$$

Title	integration of rational function of sine and cosine
Canonical name	IntegrationOfRationalFunctionOfSineAndCosine
Date of creation	2013-03-22 17:05:15
Last modified on	2013-03-22 17:05:15
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	29
Author	pahio (2872)
Entry type	Topic
Classification	msc 26A36
Synonym	universal trigonometric substitution
Related topic	GoniometricFormulae
Related topic	SubstitutionForIntegration
Related topic	WeierstrassSubstitutionFormulas
Related topic	EulersSubstitutionsForIntegration
Related topic	ErrorsCanCancelEachOtherOut
Defines	universal hyperboloc substitution