Recall that,
#sin(x-a)sin(x+a)=sin^2x-sin^2a=cos^2a-cos^2x....(ast).#
Let, #I=sqrt{sin(x-a)/sin(x+a)}dx.#
#:. I=intsqrt{sin(x-a)/sin(x+a)xxsin(x-a)/sin(x-a)}dx.#
#=intsin(x-a)/sqrt(sin(x+a)sin(x-a))dx..............................[because, (ast)],#
#=int(sinxcosa-cosxsina)/sqrt(sin(x+a)sin(x-a))dx,#
#=cosaintsinx/sqrt(sin(x+a)sin(x-a))dx#
#-sinaintcosx/sqrt(sin(x+a)sin(x-a))dx,#
#=I_1-I_2.........(1)," say, where, "#
#I_1=cosaintsinx/sqrt(sin(x+a)sin(x-a))dx,#
#=cosaintsinx/sqrt(cos^2a-cos^2x)dx,#
#=-cosaint(du)/sqrt(cos^2a-u^2)....[u=cosx rArr du=-sinxdx],#
#=-cosa*arc sin(u/cosa),#
#rArr I_1=-cosa*arc sin(cosx/cosa)......................(2).#
#"Similarly, "I_2=sinaintcosx/sqrt(sin(x+a)sin(x-a))dx,#
#=sinaintcosx/sqrt(sin^2x-sin^2a)dx.........[because, (ast),]#
#=sinaint(dv)/sqrt(v^2-sin^2a)....[v=sinx rArr dv=cosxdx],#
#=sina*ln|v+sqrt(v^2-sin^2a)|.#
#rArr I_2=sina*ln|sinx+sqrt(sin^2x-sin^2a)|............(3).#
#"By (1),(2), and,(3), then,"#
#I=-cosa*arc sin(cosx/cosa)#
#-sina*ln|sinx+sqrt(sin^2x-sin^2a)|+C#.
Enjoy Maths.!
