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Math Help - integrate

  1. #1
    mms
    mms is offline
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    integrate

    <br />
\int {\frac{{dx}}<br />
{{\sqrt {x^2 - 1} }}} <br /> <br />

    thanks!
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  2. #2
    Master Of Puppets
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    try using x = \sec(\theta)
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  3. #3
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    I dont know how to derive it but if my memory serves me right.

    <br />
\int {\frac{{dx}}<br />
{{\sqrt {x^2 - 1} }}} = \cosh^{-1}(x)+C<br />

    Just try taking the derivative of \cosh(x)
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  4. #4
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    solution brah

    Whenever you have an integral that appears to be the derivative of an inverse trig function but isn't (and cannot be solved using u-substitution), a trig substitution must be made. In your case, the integral appears to be the derivative of arcsec(x), although it is NOT. Use x = sec z.

    dx = (secztanz)dz

    I = int[secz*tanz*dz/(sqrt{(secz)^2-1})] = int[(sec z)(tan z)dz/sqrt((tan z)^2)] = int[secz dz] = int[{(sec z)^2 + secztanz}dz/{secz+tanz}]
    where the last result was found by multiplying the top and bottom by tanz + secz.

    By substitution, let u = secz+tanz. This implies the following:

    du = [secztanz+(secz)^2]dz

    Thus we have:

    I = int[du/u] = log |u|+C = log|secz + tanz|+C = log|sec(arcsecx)+tan(arcsecx))|+ C

    where log is the natural log base e and z = arcsecx.

    In general, if you have a form 1/(u^2-a^2) in the integral, use u=a*secz. If you have the form 1/(u^2 + a^2) use u = a*tanz and for 1/(a^2-u^2) use u = a*sinz.

    Now could somebody help me with my (much simpler) question posted in the calc threads?
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