#f(x)=intsinxdx-intx^2cosxdx=-cosx-I,# where,
#I=intx^2cosxdx=intuvdx", say,"# where, #u=x^2, and, v=cosx#.
But, by the Rule of Integration by Parts (ibp),
#intuvdx=uintvdx-int((du)/dxintvdx)dx#
Here, #intvdx=intcosxdx=sinx, and, (du)/dx=2x#.
#:. I=x^2sinx-int(2xsinx)dx#
#=x^2sinx-2J", where, "J=intxsinxdx#.
For, #J#, we again use ibp; this time, with #u=x, &, v=sinx#.
#:. J=x(intsinxdx)-int{d/dx(x)*intsinxdx}dx#
#=x(-cosx)-int{(1)(-cosx)}dx#,
i.e., #J=-xcosx+intcosxdx=-xcosx+sinx#
Thus, altogether, we have,
#I=x^2sinx-2J=x^2sinx-2{-xcosx+sinx}#,
or, #I=x^2sinx+2xcosx-2sinx#, so that, finally,
#f(x)=-cosx-I,#
#=-cosx-{x^2sinx+2xcosx-2sinx}#, i.e.,
#f(x)=-cosx-x^2sinx-2xcosx+2sinx+C#.
To determine #C#, we use the cond. : #f(7pi/6)=0#
#rArr -cos(7pi/6)-49pi^2/36sin(7pi/6)-7pi/3cos(7pi/6)+2sin(7pi/6)+C=0#.
#rArrsqrt3/2+49pi^2/36*1/2+7pi/3*sqrt3/2-1+C=0#
#rArr36sqrt3+49pi^2+84sqrt3pi-72+72C=0#
#rArr C=1/72(72-36sqrt3-49pi^2-84sqrt3pi)#. Therefore,
#f(x)=-cosx-x^2sinx-2xcosx+2sinx+1/72(72-36sqrt3-49pi^2-84sqrt3pi)#.
Enjoy Maths.!
