Question

simplify ln[sec(sec^-1(x/3))+tan(sec^-1(x/3))]

Answer 1

\[\int\limits_{3}^{5} 1/\sqrt(x^2-9)dx \]

Answer 2

This was where I started if that helps, I just do not know how to simplify

Answer 3

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:
\[log_e[sec(u)+tan(u)]\]
\[=log_e\left[\frac{1+sin(u)}{cos(u)}\right]\]
=

Answer 4

so before i plug u back in i should rewrite?

Answer 5

Ehhhhh yeah...Maybe you can do something regarding Taylor Series...

Answer 6

We have not learned that yet in my Cal class

Answer 7

Ohhhh well then, I suppose substitute u back in! that the most I could think of...

Answer 8

OH WAIT

Answer 9

\[\lim_{a \rightarrow 3^-}\int\limits_{a}^{5} 1/\sqrt(x^2-9)\]

Answer 10

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

Answer 11

@ybarrap I don't know why he didn't apply formula to take integral of this, but we have it, right?

Answer 12

$$
\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)



$$

Answer 13

how is tan u=sqrt(x^2-9)/3