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Question

The value of 1sin2x1cos2xdx is
  1. 12(x+tanxsecx+c)
  2. 12(xlog|sinx|+c)
  3. 12(xcosx+c)
  4. 12(log|sinx|+c)


Solution

The correct option is B 12(xlog|sinx|+c)
1sin2x1cos2xdx= (sin2x+cos2x)(2sinxcosx)(sin2x+cos2x)(cos2xsin2x)dx=(sinxcosx)22sin2xdx=12sinxcosxsinxdx=121cotx dx=12(xlog|sinx|+c)

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