Integration of a Trigonometric FunctionDate: 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/ |
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