anonymous
  • anonymous
simplify ln[sec(sec^-1(x/3))+tan(sec^-1(x/3))]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[\int\limits_{3}^{5} 1/\sqrt(x^2-9)dx \]
anonymous
  • anonymous
This was where I started if that helps, I just do not know how to simplify
anonymous
  • anonymous
Ouch. Well lets see... \[log_e[sec(sec^{-1}(x/3))+tan(sec^{-1}(x/3))]\] Let \(sec^{-1}(x/3)=u\). We can rewrite as: \[log_e[sec(u)+tan(u)]\] \[=log_e\left[\frac{1+sin(u)}{cos(u)}\right]\] =

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anonymous
  • anonymous
so before i plug u back in i should rewrite?
anonymous
  • anonymous
Ehhhhh yeah...Maybe you can do something regarding Taylor Series...
anonymous
  • anonymous
We have not learned that yet in my Cal class
anonymous
  • anonymous
Ohhhh well then, I suppose substitute u back in! that the most I could think of...
anonymous
  • anonymous
OH WAIT
anonymous
  • anonymous
\[\lim_{a \rightarrow 3^-}\int\limits_{a}^{5} 1/\sqrt(x^2-9)\]
anonymous
  • anonymous
x=3secu dx=3secu+tanu du x^2=9sec^2u u=sec^-1(x/3) this is what i have as my values so the bottom of the fraction takes care of the 3 and tan but when i integrate sec u i have this limit as a goes to 3 from the left ln[sec u +tan u] from a to 5
Loser66
  • Loser66
@ybarrap I don't know why he didn't apply formula to take integral of this, but we have it, right?
ybarrap
  • ybarrap
$$ \int_{3}^{5}\frac{1}{\sqrt{x^2-9}} dx\\\\ \text{Let x= } 3\sec u\text{ then }dx=3\sec u \tan u ~du\\\\ \int\frac{3\sec u \tan u}{\sqrt{3^2\sec^2 u-9}} du=\int\frac{3\sec u \tan u}{\sqrt{9(\sec^2u-1)}} du=\int\frac{3\sec u \tan u}{3\sqrt{\tan^2 u}} du\\ =\int\frac{\sec u \tan u}{\tan u} du=\int\sec u ~du\\ \text{Easy trick is to multiply integrand by }{\sec u + \tan u\over\sec u + \tan u}\\ \text{This gives }=\int\sec u {\sec u + \tan u\over\sec u + \tan u}~du\\=\int{{\sec^2 u + \sec u\tan u\over\sec u + \tan u}}du\\ \text{Then let }w=\sec u + \tan u\text{ then }du=\sec^2u+\sec u \tan u~du\\ \text{Integral becomes }\int \frac 1 w dw. \\ \text{Result is then }\ln|w|=\ln|\sec u + \tan u|=\ln|\frac x 3+ {\sqrt{x^2-9}\over 3} |\\ \text{Now evaluate between 3 and 5:}\\\ln\left |\frac 5 3+ {\sqrt{5^2-9}\over 3}\right |-\ln \left |\frac 3 3+ {\sqrt{3^2-9}\over 3} \right |\\ =\ln\left |\frac 5 3+ {4\over 3}\right | \\=\ln(3) $$
anonymous
  • anonymous
how is tan u=sqrt(x^2-9)/3
ybarrap
  • ybarrap
|dw:1377811880008:dw|
ybarrap
  • ybarrap
|dw:1377812023950:dw|

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