微积分学/不定积分/半角代換法

$\begin{array}{c}u=\tan \frac{x}{2}\sin x=\frac{2u}{1+u^2}\cos x=\frac{1-u^2}{1+u^2}\\ dx=\frac{2}{1+u^2}du\tan x=\frac{2u}{1-u^2}\\ \tan (\alpha \pm \beta )=\frac{\tan \alpha \pm \tan \beta }{1\mp \tan \alpha \tan \beta }\end{array}$Template:See also