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.!