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Calculus

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I don't know how to do the integral of e^(lnx^2)dx and the integral of (sin sqrtx)/(sqrtx) dx

  • Calculus - ,

    1. ∫e^(lnx^2)dx
    use the identity e^(ln(y)) = y to simplify the expression.
    2. try the substitution u=sqrt(x).

  • Calculus - ,

    Thanks for help on 1.

    on 2. if i do u=sqrt(x) my du is 1/2x^(-1/2) and that means my du is in the denominator. So it would read 2integral of sin(u)/du

  • Calculus - ,

    Nope, not on two.

    You cant solve it that way easily.

    This is difficult. Brake the sin function into its series equivalent, and integrate the series.

    http://reference.wolfram.com/mathematica/ref/SinIntegral.html

  • Calculus - ,

    forget that last answer. I am tired.

  • Calculus - ,

    For 2, almost, but not quite!
    Start with:
    u=√x
    du = (1/2)dx/√x
    dx/√x = 2du
    so
    ∫sin(√x) dx/√x
    =∫sin(u)*2du
    =-2cos(u)
    =-2cos(√x)

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