Calculus - integration

1

\int d\frac{\sin x-x\cos x}{x\sin x+\cos x}
=\frac{\sin x-x\cos x}{x\sin x+\cos x}

1

how can it solve with this method plz tell me

39

This can be done by parts also...will post soln in a while.

1

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

xsecx=1st func.,

(xcosx/(xsinx+cosx)^{2})=2nd fn

integrate by parts..........

I=xsecx(-1/(xsinx+cosx)-\int (secx+xsecxtanx)(-1)/(xsinx+cosx)dx
I=-xsecx/(xsinx+cosx)+tanx+c

integration