What is f(x) = int sinx-x^2cosx dx if f((7pi)/6) = 0 ?

What is f(x) = int (sinx-x^2cosx) dx if f((7pi)/6) = 0 ?

1 Answer
Sep 7, 2016

f(x)=-cosx-x^2sinx-2xcosx+2sinx+1/72(72-36sqrt3-49pi^2-84sqrt3pi).

Explanation:

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.!