Since, cos3x-1=(4cos^3x-3cosx)-1cos3x−1=(4cos3x−3cosx)−1, we find that,
cos3x-1cos3x−1 is a cubic polynomial in cosxcosx, and noting that the sum
of its co-efficients being 0, (cosx-1)0,(cosx−1) is a factor.
Thus, cos3x-1=4cos^3x-3cosx-1cos3x−1=4cos3x−3cosx−1,
=ul(4cos^3x-4cos^2x)+ul(4cos^2x-4cosx)+ul(cosx-1),
=4cos^2x(cosx-1)+4cosx(cosx-1)+1(cosx-1),
=(cosx-1)(4cos^2x+4cosx+1)
=(cosx-1){2(1+cos2x)+4cosx+1},
:. (cos3x-1)=(cosx-1)(2cos2x+4cosx+3).
rArr (cos3x-1)/(cosx-1)=(2cos2x+4cosx+3).
:. int(1-cos3x)/(1-cosx)dx=int(cos3x-1)/(cosx-1)dx,
=int(2cos2x+4cosx+3)dx,
=2*(sin2x)/2+4sinx+3x.
rArr int(1-cos3x)/(1-cosx)dx=sin2x+4sinx+3x+C.
Feel the Joy of Maths.!