Question

integral 1/(x^2 sqrt(x^2 - 9))

Answer 1

u may use a trig substitution :)

Answer 2

yeah I did. x = 3sec theta

Answer 3

There is a u substi along the way I dont get...

Answer 4

ehm so lets work it out

Answer 5

\[x=3 \sec \theta\]\[dx=3 \sec \theta \tan \theta d\theta\]\[\int \frac{1}{x^2 \sqrt{x^2-9}} \ dx=\int \frac{1}{9\sec^2 \theta \sqrt{3 \tan^2 \theta}} 3 \sec \theta \tan \theta \ d\theta\]

Answer 6

auchch...i have a mistake

Answer 7

am i right till there except that 3 inside the square root

Answer 8

yeah 3 tan theta

Answer 9

You will simplify all that and get 1/9 (1/sec theta)

Answer 10

so here I want to just jump right into cos theta

Answer 11

\[\int \frac{1}{9\sec \theta } d\theta=\]u r right

Answer 12

Now that I have that, what happens?

Answer 13

ok we have\[\sec \theta=\frac{1}{\cos \theta}\]

Answer 14

so ur integral becomes\[\int \frac{\cos \theta}{9} d \theta\]

Answer 15

btw the answer is sqrt(x^2 - 9)/9x

Answer 16

oh so u have difficulty in substituting back for x

Answer 17

yeah

Answer 18

so we have to find \(\sin \theta \) in terms of \(x\) knowing that \(x=3 \sec \theta\)

Answer 19

u have \[\frac{x^2}{9}=\sec^2 \theta\]\[\frac{9}{x^2}=\cos^2 \theta=1-\sin^2 \theta\]\[\sin^2 \theta=1-\frac{9}{x^2}\]does that help? :)

Answer 20

No way, thats messed up

Answer 21

How does that give me sqrt(x^2 - 9)/9x

Answer 22

just take a square root

Answer 23

Oh ok so you took the sqrt by bringing the squared to the other side

Answer 24

so i get sqrt(1-9/x^2)

Answer 25

and it'll lead us to final answer :) right?

Answer 26

yes it does :D

Answer 27

Thanks a lot! Appreciate it!

Answer 28

Just 1 question: How did you know to square both sides in x = 3 sec x --> x^2 = 9sec^2 x

Answer 29

Oh nvm because of the sin^2 + cos^2 = 1 relationship

Answer 30

ok, we know that there is a relation between sin and cos by\[\sin^2 x+\cos^2 x=1\]i just wanted to take out \(\cos^2 \theta\) somehow

Answer 31

kk

Answer 32

ah right :) good luck