How to integrate rational trigonometric functions

There is a general method to evaluate integrals of the form

$$\int^t R(\cos s,\sin s)\,ds,$$

where R is some rational function. The method is based on the substitution $u=\tan(s/2)$ and the identities

$$\sin s=2\sin(s/2)\cos(s/2)=2\tan(s/2)\cos^2(s/2)={2\tan(s/2)\over 1+\tan^2(s/2)}={2u\over 1+u^2},$$

$$\cos s=2\cos^2(s/2)-1={2\over 1+u^2}-1={1-u^2\over 1+u^2},$$

$${du\over ds}={1\over 2\cos^2(s/2)}={1+u^2\over 2}.$$

The integral then becomes

$$\int^{\tan(t/2)} R({1-u^2\over 1+u^2},{2u\over 1+u^2}){2\over 1+u^2}\,du.$$

Example:

$$\int^t {1\over\sin s}\,ds=\int^{\tan(t/2)} {1+u^2\over 2u}{2... ...^2}\,du =\int^{\tan(t/2)}{1\over u}\,du=\ln\vert\tan(t/2)\vert,$$

Note: You can use double angle formulas to show that

$$\csc(t)-\cos(t)=\tan(t/2).$$