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.

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