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calculus

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can someone show me step by step how to evaluate the integral of 2(sinx)^2 dx

I just don't understand how to do it....

  • calculus - ,

    Isn't there are identity

    sin2x=1-2sin^2 x ?

  • calculus - ,

    I don't understand what you are saying. there is no sin2x in my equation i don't think
    i have to evaluate the integral 2(sin^2x)

  • calculus - ,

    Try
    sin(x)sin(y)=(1/2)(cos(x-y)-cos(x+y))
    Put y=x and noting that cos(0)=1, you will have a simpler expression to integrate.

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