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If the Integral of (1+cos4x)/(cotx-tanx) =Acos4x+c ,then A equals

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Solution: Given  ,

                int (1+cos4x)/cotx-tanxhspace0.2cmdx=Acos4x+c

              Differentiating both side , we get

           \ Rightarrow 2cos^22x/(2cos2x/sin2x)=-4Asin4x

          \ Rightarrow                8A=-1Rightarrow A=-(1/8)

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Deependra Verma

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