Let, #I=int(1-x)sin^2(x-1)dx=-int(x-1)sin^2(x-1)dx#.
We subst. #(x-1)=y," so that, "dx=dy#.
#:. I=-intysin^2ydy#.
We appeal to the Rule of Integration by Parts :
#intuv'dy=uv-intu'vdy#, with, #u=y, &, v'=sin^2y#.
#u=y :. u'=1#.
#v'=sin^2y rArr v=intsin^2ydy=int(1-cos2y)/2dy#.
#:. v=1/2(y-1/2sin2y)#.
Utilising these, we get,
#I=-[y/2(y-1/2sin2y)-int1/2(y-1/2sin2y)dy]#,
#=-y/2(y-1/2sin2y)+1/2{y^2/2-1/2((-cos2y)/2)}#,
#=-1/2y^2+1/4ysin2y+1/4y^2+1/8cos2y#,
#=1/8cos2y+1/4ysin2y-1/4y^2#.
Replacing #y# by #(x-1)#, we have,
#I=1/8cos(2(x-1))+1/4(x-1)sin(2(x-1))-1/4(x-1)^2+C#.
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