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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