How do you find the antiderivative of #sinx/(1-cosx)#?
1 Answer
Jul 5, 2016
You can simply do a u-substitution.
Let:
#u = -cosx#
#du = sinxdx#
Therefore, the antiderivative/integral is:
#color(blue)(int sinx/(1-cosx)dx)#
#= int 1/(1+u)du#
#= ln|1+u|#
#= color(blue)(ln|1-cosx| + C)#
We know this works because:
#color(green)(d/(dx)[ln|1-cosx| + C])#
#= 1/(1-cosx)*sinx + 0#
(chain rule)
#= color(green)(sinx/(1-cosx))#
