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Inverse Trigonometric Formulas

  1. May 6, 2004 #1


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    Homework Helper
    Gold Member

    Someone has to help me here.

    1. [tex]\int \frac{dx}{\sqrt{1 - x^2}} = sin^-1 x + C[/tex]

    So the derivative of [tex]sin^-1x[/tex] is [tex]\frac{dx}{\sqrt{1 - x^2}}[/tex]?

    The -1 are exponents. It isn't working.
    Last edited: May 6, 2004
  2. jcsd
  3. May 6, 2004 #2
    The derivative of sin inverse is:

    1/[square]1 - x2[/square])
    Last edited: May 6, 2004
  4. May 6, 2004 #3


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    Do the integral with a trig substitution, x=sinu, dx=cosudu

    Use the identity sin2u+cos2u=1, or 1-sin2u=cos2u

    This way, 1-x2 becomes 1-sin2u becomes cos2u

    Your integral reduces to INTdu=u+c=sin-1+c

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