March 4th, 2015, 03:01 AM  #1 
Senior Member Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36  Integral problem
Is there any more simple method to solve $\displaystyle \int \sqrt{x^236} dx$?

March 4th, 2015, 03:14 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 06:31 AM. 
March 4th, 2015, 03:25 AM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,151 Thanks: 2390 Math Focus: Mainly analysis and algebra 
Or $6\sec u = x$

March 4th, 2015, 03:38 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, 06:00 AM  #5 
Senior Member Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36  calculus  Shortcuts for integrating $\int \sqrt{x^236} \,dx$  Mathematics Stack Exchange There're lots replies here. 
March 4th, 2015, 08:23 AM  #6 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,151 Thanks: 2390 Math Focus: Mainly analysis and algebra 
$$\newcommand{\d}[1]{\,\mathrm d #1} \begin{aligned} &\text{A standard result} & I_{1} = \int \sec x \d x &= \log{ \left\sec x + \tan x\right} + c \\[12pt] &\text{Another standard result} & I_n = \int \cos^n x \d x &= \int \cos^{n2} x \left(1\sin^2 x\right) \d x \\ && &= I_{n2}  \int \sin x \, \cos^{n2} x \sin x \d x \\ && &= I_{n2}  {1 \over n1} \sin x \cos^{n1} + {1 \over n1} \int \cos x \cos^{n1} x \d x \\ && &= I_{n2}  {1 \over n1} \sin x \cos^{n1} + {1 \over n1} I_n \\ && I_{n2} &= {1 \over n1}\sin x \cos^{n1} x + {n2 \over n1} I_n \\[8pt] && I_{3} &= {1 \over 2} \sin x \sec^2 x + {3 \over 2} \int \sec x \d x \\ && &= {3 \over 2} I_{1}  {1 \over 2}\tan x \sec x \\[12pt] && \int \sqrt{x^2  36} \d x &= 6\int \sqrt{36\sec^2 u  36} \sec u \tan u \d u \\ && &= 36 \int \sec u \left( \sec^2  1\right) \mathrm d u \\ && &= 36 \left( I_{3}  I_{1} \right) \\ && &= 36\left( {1 \over 2} I_{1}  {1 \over 2}\tan u \sec u \right) \\ && &= 18 \big( \log{ 6\left \sec u + \tan u \right}  \tan u \sec u \big) + c &\text{here the factor of 6 in the log is part of the constant of integration}\\[8pt] && \int \sqrt{x^2  36} \d x &= 18 \left( \log{ \leftx + \sqrt{x^2  36}\right}  \tfrac1{36}x \sqrt{x^2  36} \right) + c \end{aligned}$$ I wouldn't call that massively long. 
March 4th, 2015, 02:15 PM  #7 
Global Moderator Joined: Dec 2006 Posts: 18,595 Thanks: 1493 
That answer has the wrong sign.

March 4th, 2015, 02:47 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, 03:59 PM  #9 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,151 Thanks: 2390 Math Focus: Mainly analysis and algebra  
March 5th, 2015, 03:17 AM  #10 
Global Moderator Joined: Dec 2006 Posts: 18,595 Thanks: 1493 
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 $\displaystyle \frac12x\sqrt{x^2  36}  18\ln\!\leftx + \sqrt{x^2  36}\right + \text{C}\!\!$. 

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