Integration of a Trigonometric Function

Date: 06/10/99 at 11:45:54
From: Aubin
Subject: Integration of arctan(sqrt(1-x^2))

What's the method for integrating the function arctan(sqrt(1-x^2))?

Date: 06/10/99 at 13:45:46
From: Doctor Rob
Subject: Re: Integration of arctan(sqrt(1-x^2))

This is a complicated one.

First I used integration by parts with u = Arctan(sqrt(1-x^2)), 
dv = dx.

Then I substituted x = tanh(y), dx = sech^2(y) dy.

Finally I let z = tanh(y/2), so tanh(y) = 2*z/(1+z^2), sech(y) = 
(1-z^2)/(1+z^2), and dy = 2*dz/(1-z^2). This reduced the remaining 
integral to a rational function of z. (These two steps could be 
combined to substitute x = 2*z/(1+z^2), but I didn't see that this 
would be advantageous until afterwards.)

Then I factored the denominator, using

   z^4 + 1 =(z^2 + sqrt(2)*z + 1)*(z^2 - sqrt(2)*z + 1),

and used partial fractions.

The resulting integrals were easy to do, and yielded some arctangents.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/