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 August 23rd, 2012, 11:07 AM #1 Math Team     Joined: Aug 2012 From: Sana'a , Yemen Posts: 1,177 Thanks: 44 Math Focus: Theory of analytic functions Evaluate the integral $\int sec^5(x)\,\, dx$
 August 23rd, 2012, 01:30 PM #2 Global Moderator   Joined: Dec 2006 Posts: 19,294 Thanks: 1686 What have you tried?
 August 24th, 2012, 10:05 AM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,842 Thanks: 1068 Math Focus: Elementary mathematics and beyond Re: Evaluate the integral $\int\sec^5(x)\,dx\,=\,\int\sec^3(x)\sec^2(x)\,dx \\ dv\,=\,\sec^2(x)\,dx \\ v\,=\,\tan(x) \\ u\,=\,\sec^3(x) \\ du\,=\,3\sec^3(x)\tan(x)\,dx \\ \int\sec^5(x)\,dx\,=\,\sec^3(x)\tan(x)\,-\,3\int\sec^3(x)\tan^2(x)\,dx \\ =\,\sec^3(x)\tan(x)\,-\,3\int\sec^3(x)(\sec^2(x)\,-\,1)\,dx \\ =\,\sec^3(x)\tan(x)\,-\,3\int\sec^5(x)\,dx\,+\,3\int\sec^3(x)\,dx \\ 4\int\sec^5(x)\,dx\,=\,\sec^3(x)\tan(x)\,+\,3\int\ sec^3(x)\,dx \\ \text{Integral of secant cubed:} \\ \int\sec^3(x)\,dx \\ u\,=\,\sec(x),\,du\,=\,\sec(x)\tan(x) \\ dv\,=\,\sec^2(x)\,dx,\,v\,=\,\tan(x) \\ \int\sec^3(x)\,dx\,=\,\sec(x)\tan(x)\,-\,\int\sec(x)\tan^2(x)\,dx\,=\,\sec(x)\tan(x)\,-\,\int\sec^3(x)\,dx\,+\,\int\sec(x)\,dx \\ 2\int\sec^3(x)\,dx\,=\,\sec(x)\tan(x)\,+\,\ln|\sec (x)\,+\,\tan(x)| \\ \int\sec^3(x)\,dx\,=\,\frac{\sec(x)\tan(x)\,+\,\ln |\sec(x)\,+\,\tan(x)|}{2}\,+\,C \text{Now,} \\ \int\sec^5(x)\,dx\,=\,\frac{\sec^3(x)\tan(x)\,+\,\ frac32(\sec(x)\tan(x)\,+\,\ln|\sec(x)\,+\,\tan(x)| \)}{4}\,+\,C$
 August 25th, 2012, 07:10 AM #4 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Evaluate the integral Another way to do that is to immediately use $\sec(x)= \frac{1}{cos(x)}$ to write the integral as $\int \frac{dx}{cos^5(x)}$ Since that is an odd power of cosine, we can multiply both numerator and denominator by cos(x): $\int\frac{cos(x)dx}{cos^6(x)}= \int\frac{cos(x)dx}{(cos^2(x))^3}= \int\frac{cos(x)dx}{(1- sin^2(x))^3}$ Now let u= sin(x) so that du= cos(x)dx and the integral is $\int\frac{du}{(1- u^2)^3}= \int\frac{du}{(1+ u)^3(1- u)^3}$ which can be integrated by "partial fractions".
 August 25th, 2012, 11:16 AM #5 Senior Member   Joined: May 2011 Posts: 501 Thanks: 5 Re: Evaluate the integral May I ask something that has nothing to do with the integral at hand?. zaidalyafey, from where did you get the series in your avatar?. This appears to be the sum of the reciprocals of the Fibonacci numbers minus the first few terms. This does not converge to the gamma constant.
August 26th, 2012, 05:55 AM   #6
Math Team

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Re: Evaluate the integral

Quote:
 Originally Posted by galactus This does not converge to the gamma constant.

http://www.mymathforum.com/viewtopic.php?f=15&t=32679

 August 26th, 2012, 07:46 AM #7 Senior Member   Joined: May 2011 Posts: 501 Thanks: 5 Re: Evaluate the integral Ok, mathbalarka. Often times, $\psi$ is used with reference to the sum of the reciprocals of the Fibonacci numbers. Which is irrational and equal to around 3.359.... or something like that. I do not like this notation either, as $\psi$ is used for the digamma.

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