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:

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