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.!")#
