Integration by PartsDate: 04/08/99 at 00:13:10 From: Paul Auge Subject: Integration by parts Trying to solve the following integral, I wind up in a continuous loop or with everything equal to zero. INT[(sec x)^3 dx] Please help. Thanks, Paul Date: 04/08/99 at 15:27:27 From: Doctor Bruce Subject: Re: Integration by parts Hello Paul, Integration by parts will simplify this problem a little. Set it up this way: u = sec x dv = (sec x)^2 dx du = (sec x)(tan x) dx v = tan x Then you will get INT[u dv] = uv - INT [v du] INT[(sec x)^3 dx] = (sec x)(tan x) - INT[(sec x)(tan x)^2 dx]. The trick is now to use the trig identity tan^2 = sec^2 - 1 under the integral on the right side. After rearranging things a little you get 2*INT[(sec x)^3 dx] = (sec x)(tan x) + INT[(sec x) dx]. So now we have to figure out the integral of secant, rather than the integral of secant-cubed. Unfortunately, I don't know any easy way to explain how to evaluate the integral of secant. I'll just have to give you the answer: INT[(sec x) dx] = log[(1 + sin x)/cos x]. That's not very satisfying, I know. It always annoyed me when the teacher pulled a magic formula out of thin air. But there it is. You can take the derivative of the right side and see that you do get sec x, so that checks. Integration by parts would not help in integrating secant, since the answer involves logs. Remember that integration by parts only involves products. Of course, you don't know what the answer is supposed to look like! So you try integration by parts and end up chasing your tail sometimes, as you discovered. I hope this helps. - Doctor Bruce, The Math Forum http://mathforum.org/dr.math/ Date: 04/08/99 at 15:31:18 From: Doctor Rob Subject: Re: Integration by parts Thanks for writing to Ask Dr. Math! This seemingly simple integral is surprisingly hard. The proper substitution is z = tan(x/2), x = 2*arctan(z), dx = 2*dz/(1+z^2), sec(x) = (1+z^2)/(1-z^2). That transforms the integral into INT[2*(1+z^2)^2/(1-z^2)^3 dz]. This you can split using partial fractions into (1/2)*INT[1/(1-z) - 1/(1-z)^2 + 2/(1-z)^3 + 1/(1+z) - 1/(1+z)^2 + 2/(1+z)^3 dz]. Each of these terms is easily integrated. When you are done, you have to resubstitute z = tan(x/2) and simplify. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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