Question #73167
1 Answer
(sinx -xcosx)/(xsinx+cosx)+C
Explanation:
#=(-xsecx)/(xsin x+cos x)+int (sec x + xsecx tanx)/(xsinx+cosx) dx#
But
=
#1/cos^2x\cancel (xsinx +cosx)/\cancel(xsinx+cosx)#
=#sec^2x#
then
#= (-xsecx)/(xsin x+cos x)+tanx + C#
#= (xsin^2x + sin xcos x - x)/(cos x(xsin x+cos x))+C#
#=(sinx -xcosx)/(xsinx+cosx)+C#
We can verify by differentiating
#=(x^2sin^2x+\cancel(xsinxcosx)-\cancel(xsinxcosx)+x^2cos^2x)/(xsinx+cosx)^2#
=#x^2/(xsinx+cosx)^2#