What is int (1-x)sin^2(x-1) dx?

1 Answer
Apr 27, 2018

1/8cos(2(x-1))+1/4(x-1)sin(2(x-1))-1/4(x-1)^2+C.

Explanation:

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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