Evaluate the following integrals:

Assume a2 + b2sin2x = t


d(a2 + b2sin2x) = dt


2b2.sinx.cosx.dx=dt


(2sinx.cosx = sin2x)


Sin2xdx =


Put t and dt in the given equation we get


⇒


=


But t = a2 + b2sin2x


= .


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