int x^2/(xsinx+cosx)^2 dx ?

1 Answer
Jul 9, 2018

(sinx-xcosx)/(xsinx+cosx)+C.

Explanation:

Let, I=intx^2/(xsinx+cosx)^2dx,

=int{(xsecx)((xcosx)/(xsinx+cosx)^2)}dx.

We will use the following Rule of Integration by Parts (IBP) :

" IBP : "intuv'dx=uv-intu'vdx.

Prior to the Integration, let us note :

d/dx{1/(xsinx+cosx)}

=-1/(xsinx+cosx)^2*d/dx{(xsinx+cosx)},

=-1/(xsinx+cosx)^2*{(x*cosx+sinx)+(-sinx)}.

=-(xcosx)/(xsinx+cosx)^2.

rArr int(xcosx)/(xsinx+cosx)^2dx=-1/(xsinx+cosx)............(1).

Also, d/dx(xsecx)=xsecxtanx+secx,

=x*1/cosx*sinx/cosx+1/cosx.

rArr d/dx(xsecx)=(xsinx+cosx)/cos^2x............(2).

Now, in IBP, we take,

u=xsecx, and, v'=(xcosx)/(xsinx+cosx)^2.

:.u'=(xsinx+cosx)/cos^2x, &, v=-1/(xsinx+cosx).

:. I=(xsecx){-1/(xsinx+cosx)}

-int{((xsinx+cosx)/cos^2x)*(-1/(xsinx+cosx))}dx,

=-x/{cosx(xsinx+cosx)}+intsec^2xdx,

=-x/{cosx(xsinx+cosx)}+tanx,

=-x/{cosx(xsinx+cosx)}+sinx/cosx,

={-x+sinx(xsinx+cosx)}/{cosx(xsinx+cosx)},

={-x+xsin^2x+sinxcosx}/{cosx(xsinx+cosx)},

={-x(1-sin^2x)+sinxcosx}/{cosx(xsinx+cosx)},

=(-xcos^2x+sinxcosx)/{cosx(xsinx+cosx)},

={cosx(sinx-xcosx)}/{cosx(xsinx+cosx)}.

rArr I=(sinx-xcosx)/(xsinx+cosx)+C.

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