My Math Forum Challenging integral marathon!

 Calculus Calculus Math Forum

 October 6th, 2018, 05:18 PM #11 Global Moderator   Joined: Dec 2006 Posts: 20,757 Thanks: 2138 You slipped up when doing the initial substitution. Also, you forgot to include a constant of integration.
 October 6th, 2018, 07:31 PM #12 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,664 Thanks: 2644 Math Focus: Mainly analysis and algebra Apart from skipjack's comments, it looks good. But if you are going to use a substitution $x=\sec u$, why not do it on the integral $\displaystyle \int \sqrt{x^2-1} \,\mathrm dx$? Admittedly it leads to the mildly challenging $$\int \sec^3 u \, \mathrm du$$ I approached that one via \begin{align}\int \sec{(u)} \, \mathrm du &= \int \cos^2{(u)}\sec^3{(u)}\, \mathrm du \\ &= \int (1-\sin^2{(u)})\sec^3{(u)}\, \mathrm du \\ &= \int \sec^3{(u)}\,\mathrm du - \int \underbrace{\sin{(u)}}_{u} \underbrace{\sin{(u)} \cos^{-3}{(u)}}_{\mathrm dv}\, \mathrm du \end{align}
 October 6th, 2018, 08:05 PM #13 Global Moderator   Joined: Dec 2006 Posts: 20,757 Thanks: 2138 Substituting $x = \cosh u$ is simpler for that (or for integrating $x^2 /\sqrt{x^2 - 1}$) if you know about the hyperbolic functions.

 Tags challenging, integral, marathon

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Realistonia Calculus 8 February 27th, 2013 05:42 AM Etyucan Calculus 11 December 7th, 2012 10:55 AM zaidalyafey Calculus 2 November 17th, 2012 04:56 AM davedave Calculus 28 November 12th, 2012 02:36 AM galactus Complex Analysis 3 May 15th, 2012 01:39 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top