Denote, #int1/(a+bcotx)dx," by, "I." Then, "I=intsinx/(asinx+bcosx)dx.#
Evibently, #a^2+b^2ne0.#
A useful technic to solve this type of integrals, is to decompose the
Nr. as, #"Nr.="l*Dr.+m*d/dx(Dr.), where, l,m in RR....(star).#
So, let, for some #l,m in RR,#
#sinx=l(asinx+bcosx)+md/dx(asinx+bcosx)#
#:.sinx=l(asinx+bcosx)+m(acosx-bsinx), i.e., #
#1sinx+0cosx=(la-mb)sinx+(lb+ma)cosx.#
#:. la-mb=1, &, lb+ma=0.# Solving these for #l,m;# we get,
#l=a/(a^2+b^2), m=-b/(a^2+b^2)...........(star')#
Now, #(star) rArr sinx/(asinx+bcosx)=l+{md/dx(asinx+bcosx)}/(asinx+bcosx).#
#:. I=int[l+{md/dx(asinx+bcosx)}/(asinx+bcosx)]dx,#
#=l*int1dx+m*int{(d/dx(asinx+bcosx))/(asinx+bcosx)}dx,#
#=lx+mln|(asinx+bcosx)|.#
Finally, using #(star'),# we have,
#I=a/(a^2+b^2)*x-b/(a^2+b^2)*ln|(asinx+bcosx)|, or,#
#I=1/(a^2+b^2){ax-bln|(asinx+bcosx)|}+C.#
Foot Note :
Observe that, the above Solution is based on the valid
condition that #a^2+b^2 ne0.#
I hope that the Questioner will work out the Solution for
#a=0 ( &, :., bne0), and, b=0; (ane0.)#
Enjoy Maths.!