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.
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Integration query. watch
- Thread Starter
- 21-11-2005 17:52
- 21-11-2005 18:26
Suppose both f(x) and g(x) differentiate to give you , but and
Therefore we have by construction. Integrate to get where K is a constant. Since we know g(x) we get .
Whatever expression you derive will be but just with a constant added on.
If you mean "Is there a representation of 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 give or take a constant.
does have a rather pleasant Maclaurin series, namely
This gives the exceptionally elegant series for if you put in x=1 as
Unfortunately the rate of convergence is shockingly terrible, needing thousands of terms to get only 3 or 4 decimal places.