How do you find the antiderivative of #1/(1-cosx)#?

Question

10768 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-04-17T17:58:57

#int1/(1-cos(x)) dx#

First, let's do some trigonometric transformation :

Remember that #=>sin^2(x) = 1/2(1-cos(2x))#

(From #sin(a)*sin(b)# formula with #a = b#)

#=>2sin^2(x) = 1-cos(2x)#

#=>2sin^2(1/2x) = 1-cos(x)#

So we can write :

#int1/2*1/(sin^2(1/2x))dx#

Divise numerator and denominator by #cos^2(1/2x)#

#int1/2*(1/cos^2(1/2x))/(sin^2(1/2x)/cos^2(1/2x))dx =int1/2*(1/cos^2(1/2x))/tan^2(1/2x)dx#

let's #t = tan(1/2x)#

#=> dt = 1/2*1/cos^2(1/2x)#

Just have a look we have #dt# in the integral !

It's perfect :

#int1/t^2dt = [-1/t]#

Substitute back for t = tan(1/2x)

=> #-1/tan(1/2x)+C#