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

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#.

#color(blue)("Enjoy Maths.!")#