tan(\(\frac{x}{2}\))=tとおくと　
dt=(\(\frac{1}{2}\))*{\(\frac{1}{c}\)os(\(\frac{x}{2}\))}^2dx
　=(\(\frac{1}{2}\))*{1+{tan(\(\frac{x}{2}\))}^2}dx
　=(\(\frac{1}{2}\))*(1+\(t^{2}\))dx

よって
dx={2/(\(t^{2}\)+1)}dt

(1-\(t^{2}\))/(1+\(t^{2}\))
＝(1-{sin(\(\frac{x}{2}\))/cos(\(\frac{x}{2}\))}^2)/(1+{sin(\(\frac{x}{2}\))/cos(\(\frac{x}{2}\))}^2)
＝({cos(\(\frac{x}{2}\))}^2-{sin(\(\frac{x}{2}\))}^2)/({cos(\(\frac{x}{2}\))}^2+{sin(\(\frac{x}{2}\))}^2)
＝cosx

だから

∫1/(1-cosx)dx
＝∫{1/(1-{(1-\(t^{2}\))/(1+\(t^{2}\))})}*{2/(\(t^{2}\)+1)}dt
＝∫{(1+\(t^{2}\))/(1+\(t^{2}\))-(1-\(t^{2}\))}*{2/(\(t^{2}\)+1)}dt
＝∫{(1+\(t^{2}\))/(2\(t^{2}\))}*{2/(\(t^{2}\)+1)}dt
＝∫(1/\(t^{2}\))dt
＝-\(\frac{1}{t}\)+C
＝-1/{tan(\(\frac{x}{2}\))}+C

Cは積分定数

1/(1-cosx)
=(1+cosx)/(1-(cosx\()^{2}\))
=(1+cosx)/(sinx\()^{2}\)

∫dx/(1-cosx)
=∫dx/(sinx\()^{2}\)+∫cosx/(sinx\()^{2}\)dx
=-cotx-\(\frac{1}{s}\)inx+C