Though the Question is not clear, but we solve it by taking #2# possibilities :
Alternative I :-
#int(sin^2x-cos^2x)/sinx dx=int{(sin^2x-(1-sin^2x)}/sinxdx#
#=int(2sin^2x-1)/sinx dx=int{(2sin^2x)/sinx-1/sinx}dx#
#=2intsinxdx-intcscxdx#
#=2(-cosx)-ln|cscx-cotx|#
#=-{2cosx+ln|cscx-cotx|+C.#
Alternative II :-
#int{sin^2x-cos^2x/sinx}dx#
We use the Identity # : sin^2x=(1-cos2x)/2#.
#:. int{sin^2x-cos^2x/sinx}dx#
#=int(1-cos2x)/2dx-int(1-sin^2x)/sinxdx#
#1/2(x-sin(2x)/2)-int(1/sinx-sinx)dx#
#=1/4(2x-sin2x)-intcscxdx+intsinxdx#
#=1/4(2x-sin2x)-ln|cscx-cotx|-cosx+C#
Enjoy Maths.!
