Question

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

Answer 1

ick

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

hmm

Answer 12

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

Answer 13

Oh, I see, I see.

Answer 14

Are you able to continue it?

Answer 15

But looks like i made a mistake somewhere.,

Answer 16

hold the phone

Answer 17

I'm just writing things out.

Answer 18

do a little trig simplifying first

Answer 19

i think i should do it on paper lol

something looks wrong

Answer 20

Yes, something does look wrong.

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

Yeah I got that. Thanks a lot!

Answer 28

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