integrate [sin(x-a)/sin(x+a)]1/2 w.r.t x

Hi Suchita,

This becomes, (sinxcosa - cosxsina / sinxcosa + cosxsina)^1/2

ie (tanx - tana / tanx + tana)^1/2

make the substitution tana + tanx = t²

So that the integral becomes 2(-2tana + t²)^1/2/{1+{t²-tana)²}

Take t² common form both the Nr and Dr, and make the substitution tana/t² = y, so that 1/t³ dt = dy/(-2tana)

In the denominator you will have terms like, {y²/tan²a + (1-y)²}, which is a quadratic.

You will get a standard integral of the form ∫(√linear / Quadratic)

Which is integrated by the standard procedure.

Hope that helps.

All the best,

Regards,

Ashwin (IIT Madras).