integration under root (sin2x -1)

Dear Student,
Please find below the solution to the asked query:

I think it should be 1-sin2x because as -1sin2x1, so sin2x-1 will only be defined whensin2x=1, so fx=sin2x-1 will always be 0. 1-1=0So, we have,I=1-sin2xdx=sin2+cos2x-2sinxcosxdx=sinx-cosx2dx=sinx-cosxdx=-cosx-sinx +C1-sin2xdx=-sinx+cosx+C

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