#int (1-cos 3x)/(1-cos x)# ?

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Answer 1 · 2018-05-31T05:25:39

# sin2x+4sinx+3x+C#.

Since, #cos3x-1=(4cos^3x-3cosx)-1#, we find that,

#cos3x-1# is a cubic polynomial in #cosx#, and noting that the sum

of its co-efficients being #0, (cosx-1)# is a factor.

Thus, #cos3x-1=4cos^3x-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#.

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