Integrate. sqrt(x^2-9)/x^3
ick
\[\int\limits\frac{\sqrt{x^{2}-9}}{x^{3}} \] Trig-sub. \[x=asec\theta => x=3\sec\theta \]
try \(u=3\sec(\theta)\)
\[dx = 3\sec\theta \tan\theta d\theta \]
actually it is not really that bad, because you will end up cancelling a bunch of stuff mimi is on the case
I got up to \[\int\limits_{}^{}\tan^2 \theta/3\sec^2\theta d \theta\]
After that I'm not sure. I looked at changing tan^2=1+sec^2
hmm \[\int\limits\frac{\sqrt{x^{2}-9}}{x^{3}} => \int\limits\frac{\sqrt{(3\sec\theta)^{2}-9}}{(3\sec\theta)^{3}} *3\sec\theta \tan\theta d\theta \] \[=> \int\limits\frac{\sqrt{9\sec^{2}\theta-9}}{27\sec^{3}\theta} *3\sec\theta \tan\theta d\theta =>\int\limits\frac{\sqrt{9(\sec^{2}\theta-1)}}{27\sec^{3}\theta} *3\sec\theta \tan\theta d\theta => \int\limits\frac{\sqrt{9\tan^{2}\theta}}{27\sec^{3}\theta} *3\sec\theta \tan\theta d\theta \]
\[=> \int\limits\frac{\sqrt{9\tan^{2}\theta}}{27\sec^{3}\theta} *3\sec\theta \tan\theta d\theta \]
\[=> \int\limits\frac{3\tan\theta}{27\sec^{3}\theta} *3\sec\theta \tan\theta d\theta \]
hmm
there is something wrong weith this site it keeps on freezing!! \[=> \int\limits\frac{9\tan^{2}\theta \sec\theta}{27\sec^{3}\theta} => \int\limits\frac{9(\sec^{2}\theta-1)\sec\theta}{27\sec^{3}\theta} => \int\limits\frac{(9\sec^{2}-9)\sec\theta}{27\sec^{3}\theta} => \int\limits\frac{9\sec^{3}\theta-9\sec\theta}{27\sec^{3}\theta} \]
Oh, I see, I see.
Are you able to continue it?
But looks like i made a mistake somewhere.,
hold the phone
I'm just writing things out.
do a little trig simplifying first
i think i should do it on paper lol something looks wrong
Yes, something does look wrong.
what you neglected to do was simplify \[\frac{\tan^2(\theta)\sec(\theta)}{\sec^2(\theta)}=\sin^2(\theta)\]
that is why i wrote above that it was not that bad, a bunch of stuff cancels
im beginning to wonder if satellite can integrate mentally...
you get \[\frac{1}{3}\int\sin^2(\theta)d\theta\] and integral you can probably look up
no but i can sort of reduce fractions mentally, at least sometimes
To integrate: sin^{2}theta wriite it as: \[\sin^{2}\theta = \frac{1-\cos2\theta}{2} \]
Yeah I got that. Thanks a lot!
\[\frac{\tan^2(x)\sec(x)}{\sec^3(x)}=\frac{\tan^2(x)}{\sec^2(x)}=\frac{\sin^(x)}{\cos^2(x)}\times \cos^2(x)=\sin^2(x)\]
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