Integration of : #int(cosx-cos2x)/(1-cosx) dx# ?

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Answer 1 · 2018-03-15T13:41:43

The answer is #=2sinx+x+C#

We need

#cos2x=2cos^2x-1#

Therefore,

#int((cosx-cos2x)dx)/(1-cosx)=int((cos2x-cosx)dx)/(cosx-1)#

#=int((2cos^2x-1-cosx)dx)/(cosx-1)#

#=int((2cosx+1)(cosx-1)dx)/(cosx-1)#

#=int(2cosx+1)dx#

#=2sinx+x+C#

Answer 2 · 2018-03-15T14:38:11

# x+2sinx+C#.

Here is another approach to find the integral :

#(cosx-cos2x)/(1-cosx)#,

#={-2sin((x+2x)/2)sin((x-2x)/2)}/(2sin^2(x/2)#,

#=sin(3/2x)/sin(1/2x)#,

#={3sin(1/2x)-4sin^3(1/2x)}/sin(1/2x)#,

#=3-4sin^2(1/2x)#,

#=3-2{2sin^2(1/2x)}#,

#=3-2(1-cosx)#,

#=1+2cosx#.

# rArr int(cosx-cos2x)/(1-cosx)dx#,

#=int(1+2cosx)dx#,

#=x+2sinx+C,# as respected Narad T. has derived!