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Integration problem

  • Thread starter temaire
  • Start date
  • #1
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Homework Statement



Find the integral of [tex]\int\frac{5dx}{\sqrt{25x^2 -9}}, x > \frac{3}{5}[/tex]



The Attempt at a Solution



First, I made x = 3/5 secx, and dx = 3/5 secxtanxdx

[tex]\int\frac{3secxtanxdx}{5\sqrt{(9/25)(sec^2x -1)}}[/tex]

[tex]\int\frac{secxtanxdx}{tanx}[/tex]

[tex]\int secxdx[/tex]

[tex]ln|secx + tanx| + C[/tex]

[tex]ln|\frac{5x}{3} + \frac{5\sqrt{x^2 - \frac{9}{25}}}{3}| + C[/tex]

The final step is my answer. However, when I try to integrate using the wolfram integration calculator, I get [tex]ln|2(\sqrt{25x^2 - 9} + 5x) + C[/tex]

Where did I go wrong?
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,230
31
Well I can't find an error in what you did but what I can say is that their answer can be reduced to


ln2+ln|5x+√(25x2-9)|+C=ln|5x+√(25x2-9)|+A


and your answer can be written as

ln(1/3)+ln|5x+√(25x2-9)| = ln|5x+√(25x2-9)|+B

So I would say that they are the same in essence.
 
  • #3
33,481
5,171
temaire said:
First, I made x = 3/5 secx, and dx = 3/5 secxtanxdx
It's not a good idea to have a substitution variable with the same name as the variable it is a substitution for.
 

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