How do you Integrate #cotxdx# by using substitution?

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Answer 1 · 2015-02-26T15:06:08

Some basic identities that we'll need:
#cos(x) = (cos(x))/(sin(x))#

#int (1/w) dw = ln(|w|) + C#

#( d sin(x))/(dx) = cos(x)#

O.K. here we go:
#int (cot(x)) dx = int ( (cos(x))/(sin(x)) ) dx#

If we let
#w = sin(x)# then #(dw)/(dx) = cos(x)#
so
#int (cos(x))/(sin (x)) dx = int ( ((dw)/(dx))/w) dx#

=#int 1/w dw#

#= ln(|w|) + C#

#= ln(|sin(x)|) + C#