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    Is it possible to inegrate the derivatives of the inverse trig functions to obtain a function of x that doesn't actually include the trigonometric terms?

    For instance, the derivative of tan-1x is 1/(1+x²). Is it possible to then inegrate 1/(1+x²) using some substitution to obtain a function of x excluding tan-1x?

    Because from this you could then find the inverse function which would give you tanx in terms of x and excluding tan. Just a thought.

    Suppose both f(x) and g(x) differentiate to give you \frac{1}{1+x^{2}}, but f(x) \not= g(x) and g(x) = \tan^{-1}(x)

    Therefore we have \frac{df}{dx} = \frac{dg}{dx} by construction. Integrate to get f(x) = g(x) + K where K is a constant. Since we know g(x) we get f(x) = \tan^{-1}(x) + K.

    Whatever expression you derive will be \tan^{-1}(x) but just with a constant added on.

    If you mean "Is there a representation of tan^{-1}(x) which doesn't actually involve tan?" then yes there is, but it involves complex logorithms, and isn't particularly nice, and as I've just proved, it will be \tan^{-1}(x) give or take a constant.

    \tan^{-1}(x) does have a rather pleasant Maclaurin series, namely

    \tan^{-1}(x) = x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - \frac{x^{7}}{7} + ...

    This gives the exceptionally elegant series for \pi if you put in x=1 as

    \frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + ...

    Unfortunately the rate of convergence is shockingly terrible, needing thousands of terms to get only 3 or 4 decimal places.
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Updated: November 21, 2005

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