Calc II integration by trig sub

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SUMMARY

The integral from \(\sqrt{3}\) to \(3\) of \(\frac{\sqrt{x^2 + 9}}{x}dx\) was evaluated using the substitution \(x = 3\tan(u)\). The user derived a solution involving secant functions, specifically \(\sec(\pi/4) - \sec(\pi/6)\), but found discrepancies with the textbook answer, which is \(\frac{1}{6}\ln((3-2\sqrt{2})(7+4\sqrt{2}))\). The discussion highlights the importance of verifying each step in the substitution process to ensure accuracy in integral evaluation.

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  • Understanding of integral calculus, specifically techniques of integration.
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  • Review trigonometric substitution techniques in integral calculus.
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Students studying calculus, particularly those focusing on integration techniques, as well as educators looking for examples of trigonometric substitution in practice.

Rock32
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Homework Statement



Find the integral from 31/2 to 3 of 1/(x) ((x2 +9))1/2


I tried to substitute x= 3tanu, and I get down eventually to a solution that is (sec(pi/4) - sec(pi/6))

I realize integrals can have many forms, but the form in the back of my book isn't even close to sec(arctan(3/3)-sec(arctan(31/2/3). My answer doesn't match the books. Maybe its a domain problem?

The books answer is: 1/6ln((3-2*21/2)(7+4*31/2

Sorry, I'm exhausted, and I do not have the energy to type out how I got from the tangent substitution to my final value.

Any help would be great.
 
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Is this your integral?
\int_{\sqrt{3}}^3 \frac{\sqrt{x^2 + 9}}{x}dx
Your substitution looks to be the correct one, but if you're not getting the same answer as in your text, possibly you made a mistake after doing the substitution. It's also possible, but less probable, that the book's answer is wrong.
 

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