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Mathematics 13 Online
OpenStudy (anonymous):

Integrate. sqrt(x^2-9)/x^3

OpenStudy (anonymous):

ick

OpenStudy (mimi_x3):

\[\int\limits\frac{\sqrt{x^{2}-9}}{x^{3}} \] Trig-sub. \[x=asec\theta => x=3\sec\theta \]

OpenStudy (anonymous):

try \(u=3\sec(\theta)\)

OpenStudy (mimi_x3):

\[dx = 3\sec\theta \tan\theta d\theta \]

OpenStudy (anonymous):

actually it is not really that bad, because you will end up cancelling a bunch of stuff mimi is on the case

OpenStudy (anonymous):

I got up to \[\int\limits_{}^{}\tan^2 \theta/3\sec^2\theta d \theta\]

OpenStudy (anonymous):

After that I'm not sure. I looked at changing tan^2=1+sec^2

OpenStudy (mimi_x3):

hmm \[\int\limits\frac{\sqrt{x^{2}-9}}{x^{3}} => \int\limits\frac{\sqrt{(3\sec\theta)^{2}-9}}{(3\sec\theta)^{3}} *3\sec\theta \tan\theta d\theta \] \[=> \int\limits\frac{\sqrt{9\sec^{2}\theta-9}}{27\sec^{3}\theta} *3\sec\theta \tan\theta d\theta =>\int\limits\frac{\sqrt{9(\sec^{2}\theta-1)}}{27\sec^{3}\theta} *3\sec\theta \tan\theta d\theta => \int\limits\frac{\sqrt{9\tan^{2}\theta}}{27\sec^{3}\theta} *3\sec\theta \tan\theta d\theta \]

OpenStudy (mimi_x3):

\[=> \int\limits\frac{\sqrt{9\tan^{2}\theta}}{27\sec^{3}\theta} *3\sec\theta \tan\theta d\theta \]

OpenStudy (mimi_x3):

\[=> \int\limits\frac{3\tan\theta}{27\sec^{3}\theta} *3\sec\theta \tan\theta d\theta \]

OpenStudy (lgbasallote):

hmm

OpenStudy (mimi_x3):

there is something wrong weith this site it keeps on freezing!! \[=> \int\limits\frac{9\tan^{2}\theta \sec\theta}{27\sec^{3}\theta} => \int\limits\frac{9(\sec^{2}\theta-1)\sec\theta}{27\sec^{3}\theta} => \int\limits\frac{(9\sec^{2}-9)\sec\theta}{27\sec^{3}\theta} => \int\limits\frac{9\sec^{3}\theta-9\sec\theta}{27\sec^{3}\theta} \]

OpenStudy (anonymous):

Oh, I see, I see.

OpenStudy (mimi_x3):

Are you able to continue it?

OpenStudy (mimi_x3):

But looks like i made a mistake somewhere.,

OpenStudy (anonymous):

hold the phone

OpenStudy (anonymous):

I'm just writing things out.

OpenStudy (anonymous):

do a little trig simplifying first

OpenStudy (mimi_x3):

i think i should do it on paper lol something looks wrong

OpenStudy (anonymous):

Yes, something does look wrong.

OpenStudy (anonymous):

what you neglected to do was simplify \[\frac{\tan^2(\theta)\sec(\theta)}{\sec^2(\theta)}=\sin^2(\theta)\]

OpenStudy (anonymous):

that is why i wrote above that it was not that bad, a bunch of stuff cancels

OpenStudy (lgbasallote):

im beginning to wonder if satellite can integrate mentally...

OpenStudy (anonymous):

you get \[\frac{1}{3}\int\sin^2(\theta)d\theta\] and integral you can probably look up

OpenStudy (anonymous):

no but i can sort of reduce fractions mentally, at least sometimes

OpenStudy (mimi_x3):

To integrate: sin^{2}theta wriite it as: \[\sin^{2}\theta = \frac{1-\cos2\theta}{2} \]

OpenStudy (anonymous):

Yeah I got that. Thanks a lot!

OpenStudy (anonymous):

\[\frac{\tan^2(x)\sec(x)}{\sec^3(x)}=\frac{\tan^2(x)}{\sec^2(x)}=\frac{\sin^(x)}{\cos^2(x)}\times \cos^2(x)=\sin^2(x)\]

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