Answer :

Assume acos2x + bsin2x = t


d(acos2x + bsin2x) = dt


( - 2acosx.sinx + 2bsinx.cosx)dx = dt


(bsin2x - asin2x)dx=dt


(b - a)sin2xdx=dt


Sin2xdx =


Put t and dt in given equation we get



= ln|t| + c.


But t = acos2x + bsin2x


= ln| acos2x + bsin2x | + c.


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