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.!
