October 6th, 2018, 06:18 PM  #11 
Global Moderator Joined: Dec 2006 Posts: 19,951 Thanks: 1842 
You slipped up when doing the initial substitution. Also, you forgot to include a constant of integration. 
October 6th, 2018, 08:31 PM  #12 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,508 Thanks: 2513 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 $\displaystyle \int \sqrt{x^21} \,\mathrm dx$? 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, 09:05 PM  #13 
Global Moderator Joined: Dec 2006 Posts: 19,951 Thanks: 1842 
Substituting $x = \cosh u$ is simpler for that (or for integrating $x^2 /\sqrt{x^2  1}$) if you know about the hyperbolic functions.


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challenging, integral, marathon 
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