My Math Forum Integral problem

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 March 4th, 2015, 02:01 AM #1 Senior Member   Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 Integral problem Is there any more simple method to solve $\displaystyle \int \sqrt{x^2-36} dx$?
 March 4th, 2015, 02:14 AM #2 Senior Member     Joined: Dec 2013 From: some subspace Posts: 212 Thanks: 72 Math Focus: real analysis, vector analysis, numerical analysis, discrete mathematics Substitute e.g. $\displaystyle x = 6\cosh t$. Last edited by skipjack; March 4th, 2015 at 05:31 AM.
 March 4th, 2015, 02:25 AM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,403 Thanks: 2477 Math Focus: Mainly analysis and algebra Or $6\sec u = x$
 March 4th, 2015, 02:38 AM #4 Senior Member   Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 Fysmat, can you show the solution ? Archie, I solved using your way, but it's a bit long. Can you show to me also? Thanks
 March 4th, 2015, 05:00 AM #5 Senior Member   Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 calculus - Shortcuts for integrating $\int \sqrt{x^2-36} \,dx$ - Mathematics Stack Exchange There're lots replies here. Thanks from fysmat
 March 4th, 2015, 07:23 AM #6 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,403 Thanks: 2477 Math Focus: Mainly analysis and algebra \newcommand{\d}[1]{\,\mathrm d #1} \begin{aligned} &\text{A standard result} & I_{-1} = \int \sec x \d x &= \log{ \left|\sec x + \tan x\right|} + c \\[12pt] &\text{Another standard result} & I_n = \int \cos^n x \d x &= \int \cos^{n-2} x \left(1-\sin^2 x\right) \d x \\ && &= I_{n-2} - \int \sin x \, \cos^{n-2} x \sin x \d x \\ && &= I_{n-2} - {1 \over n-1} \sin x \cos^{n-1} + {1 \over n-1} \int \cos x \cos^{n-1} x \d x \\ && &= I_{n-2} - {1 \over n-1} \sin x \cos^{n-1} + {1 \over n-1} I_n \\ && I_{n-2} &= {1 \over n-1}\sin x \cos^{n-1} x + {n-2 \over n-1} I_n \\[8pt] && I_{-3} &= {1 \over -2} \sin x \sec^2 x + {-3 \over -2} \int \sec x \d x \\ && &= {3 \over 2} I_{-1} - {1 \over 2}\tan x \sec x \\[12pt] && \int \sqrt{x^2 - 36} \d x &= 6\int \sqrt{36\sec^2 u - 36} \sec u \tan u \d u \\ && &= 36 \int \sec u \left( \sec^2 - 1\right) \mathrm d u \\ && &= 36 \left( I_{-3} - I_{-1} \right) \\ && &= 36\left( {1 \over 2} I_{-1} - {1 \over 2}\tan u \sec u \right) \\ && &= 18 \big( \log{ 6\left| \sec u + \tan u \right|} - \tan u \sec u \big) + c &\text{here the factor of 6 in the log is part of the constant of integration}\\[8pt] && \int \sqrt{x^2 - 36} \d x &= 18 \left( \log{ \left|x + \sqrt{x^2 - 36}\right|} - \tfrac1{36}x \sqrt{x^2 - 36} \right) + c \end{aligned} I wouldn't call that massively long.
 March 4th, 2015, 01:15 PM #7 Global Moderator   Joined: Dec 2006 Posts: 19,547 Thanks: 1752 That answer has the wrong sign.
 March 4th, 2015, 01:47 PM #8 Senior Member   Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 In another site. Someone said that it can be solved by Euler substitution. Can anyone show it? Thanks.
March 4th, 2015, 02:59 PM   #9
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Quote:
 Originally Posted by skipjack That answer has the wrong sign.
Yes. The result of some careless rearranging in the reduction formula which, as I pointed out, could be quoted as a standard result.

 March 5th, 2015, 02:17 AM #10 Global Moderator   Joined: Dec 2006 Posts: 19,547 Thanks: 1752 Let $\displaystyle t = x + \sqrt{x^2 - 36}$, so that $\displaystyle x = t/2 + 18/t$ and $\displaystyle dx = (1/2 - 18/t^2)dt$, then $\displaystyle \!\int\!\sqrt{x^2 - 36}\,dx$ becomes $\displaystyle \int(t/2 - 18/t)(1/2 - 18/t^2)dt = \!\int(t/4 + 324/t^3 - 18/t)dt = t^2/8 - 162/t^2 - 18\ln\!|t| + \text{C}\!\!\!\!$, which is $\displaystyle \frac12x\sqrt{x^2 - 36} - 18\ln\!\left|x + \sqrt{x^2 - 36}\right| + \text{C}\!\!$. Thanks from fysmat

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