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March 4th, 2015, 03:01 AM   #1
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Integral problem

Is there any more simple method to solve $\displaystyle \int \sqrt{x^2-36} dx$?
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March 4th, 2015, 03:14 AM   #2
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Substitute e.g. $\displaystyle x = 6\cosh t$.

Last edited by skipjack; March 4th, 2015 at 06:31 AM.
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March 4th, 2015, 03:25 AM   #3
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Or $6\sec u = x$
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March 4th, 2015, 03:38 AM   #4
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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
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March 4th, 2015, 06:00 AM   #5
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calculus - Shortcuts for integrating $\int \sqrt{x^2-36} \,dx$ - Mathematics Stack Exchange

There're lots replies here.
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March 4th, 2015, 08:23 AM   #6
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$$\newcommand{\d}[1]{\,\mathrm d #1} \begin{aligned} &\text{A standard result} & I_{-1} = \int \sec x \d x &= \log{ \left|\sec x + \tan x\right|} + c \\[12pt] &\text{Another standard result} & I_n = \int \cos^n x \d x &= \int \cos^{n-2} x \left(1-\sin^2 x\right) \d x \\ && &= I_{n-2} - \int \sin x \, \cos^{n-2} x \sin x \d x \\ && &= I_{n-2} - {1 \over n-1} \sin x \cos^{n-1} + {1 \over n-1} \int \cos x \cos^{n-1} x \d x \\ && &= I_{n-2} - {1 \over n-1} \sin x \cos^{n-1} + {1 \over n-1} I_n \\ && I_{n-2} &= {1 \over n-1}\sin x \cos^{n-1} x + {n-2 \over n-1} I_n \\[8pt]
&& I_{-3} &= {1 \over -2} \sin x \sec^2 x + {-3 \over -2} \int \sec x \d x \\ && &= {3 \over 2} I_{-1} - {1 \over 2}\tan x \sec x \\[12pt]
&& \int \sqrt{x^2 - 36} \d x &= 6\int \sqrt{36\sec^2 u - 36} \sec u \tan u \d u \\ && &= 36 \int \sec u \left( \sec^2 - 1\right) \mathrm d u \\ && &= 36 \left( I_{-3} - I_{-1} \right) \\ && &= 36\left( {1 \over 2} I_{-1} - {1 \over 2}\tan u \sec u \right) \\ && &= 18 \big( \log{ 6\left| \sec u + \tan u \right|} - \tan u \sec u \big) + c &\text{here the factor of 6 in the log is part of the constant of integration}\\[8pt] && \int \sqrt{x^2 - 36} \d x &= 18 \left( \log{ \left|x + \sqrt{x^2 - 36}\right|} - \tfrac1{36}x \sqrt{x^2 - 36} \right) + c \end{aligned}$$
I wouldn't call that massively long.
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March 4th, 2015, 02:15 PM   #7
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That answer has the wrong sign.
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March 4th, 2015, 02:47 PM   #8
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In another site. Someone said that it can be solved by Euler substitution. Can anyone show it? Thanks.
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March 4th, 2015, 03:59 PM   #9
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Quote:
Originally Posted by skipjack View Post
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.
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March 5th, 2015, 03:17 AM   #10
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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

$\displaystyle \frac12x\sqrt{x^2 - 36} - 18\ln\!\left|x + \sqrt{x^2 - 36}\right| + \text{C}\!\!$.
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