[SPLIT] int [ 1 / sqrt(x^2 - 9) ] dx

[warwick]

\(\displaystyle \L\\\int{dx/(x^{2}-9)^{1/2}}dx\)

x = 3 sec (theta)
3 sec (theta) tan (theta) d(theta) = dx
theta = arcsec (x/3)

\(\displaystyle \L\\\int{1/(9tan^{2} (theta))^{1/2}} 3 sec (theta) tan (theta) d(theta)\)

\(\displaystyle \L\\\int{sec (theta)}d(theta)\)

ln absolute value of (x/3) + [(x^{2}-9)^(1/2)]/3

 

[galactus]

\(\displaystyle \L\\\int{sec({\theta})d{\theta}=ln(sec{\theta}+tan{\theta})\)


\(\displaystyle \L\\sec{\theta})=\frac{x}{3}\)

\(\displaystyle \L\\tan{\theta}=\frac{\sqrt{x^{2}-9}}{3}\)

So,

\(\displaystyle \L\\ln(\frac{x}{3}+\frac{\sqrt{x^{2}-9}}{3})=ln(\frac{\sqrt{x^{2}-9}+x}{3})=ln(\sqrt{x^{2}-9}+x)-ln(3)\)

\(\displaystyle \L\\ln(\sqrt{x^{2}-9}+x)+C\)

 

[warwick]

\(\displaystyle \L\\\int{sec({\theta})d{\theta}=ln(sec{\theta}+tan{\theta})\)


\(\displaystyle \L\\sec{\theta})=\frac{x}{3}\)

\(\displaystyle \L\\tan{\theta}=\frac{\sqrt{x^{2}-9}}{3}\)

So,

\(\displaystyle \L\\ln(\frac{x}{3}+\frac{\sqrt{x^{2}-9}}{3})=ln(\frac{\sqrt{x^{2}-9}+x}{3})=ln(\sqrt{x^{2}-9}+x)-ln(3)\)

\(\displaystyle \L\\ln(\sqrt{x^{2}-9}+x)+C\)

Yeah, I just wanted to make sure that it worked out like that. Thanks. I might have a couple more before the night is over.