The integral is determined using the method of integration by parts :
color(red)(intu(x)dv(x)=u(x)v(x)-intv(x)du(x))
Take into consideration the integral of the trigonometric function
color(red)(intsin(x-1)dx=-cos(x-1))
Let us compute the given integral:
consider
color(blue)(u(x)=x^2rArrd(u(x))=2xdx)
then,
color(brown)(dv(x)=sin(x-1)rArrv(x)=-cos(x-1))
intx^2sin(x-1)dx integral 1
=color(red)(u(x)*v(x)-intdu(x)v(x))
=color(blue)x^2*color(brown)(-cos(x-1))-intcolor(blue)(2xdx)color(brown)((-cos(x-1))
=-x^2cos(x-1)+2color(purple)(intxcos(x-1)dx)
color(purple)(intxcos(x-1)dx)
let a(x)=xrArrda(x)=dx
then db(x)=cos(x-1)rArrb(x)=sin(x-1)
color(purple)(intxcos(x-1)dx)
=color(red)(a(x)*b(x)-intb(x)da(x))
=xsin(x-1)-intsin(x-1)dx
=xsin(x-1)-intd(-cos(x-1))
=xsin(x-1)+intdcos(x-1)
color(purple)(=xsin(x-1)+cos(x-1))
Let us substitute color(purple)(intxcos(x-1)dx)
in integral 1
intx^2sin(x-1)dx
=-x^2cos(x-1)+2color(purple)(intxcos(x-1)dx)
=-x^2cos(x-1)+2(xsin(x-1)+cos(x-1))
=-x^2cos(x-1)+2xsin(x-1)+2cos(x-1)
Therefore,
intx^2sin(x-1)dx
=-x^2cos(x-1)+2xsin(x-1)+2cos(x-1)
