Question

find the integral of 1/sqrt(x^2-9)

Answer 1

thank you!

Answer 2

when you wrote 3cos^2u in this step: "3 sqrt ( 1 - sin^2u) = 3 cos^2 u"... cos is not supposed to be squared right?

Answer 3

That's why I'm double checking it now! I'm supposedly work out it on paper first.

Answer 4

= - 9 Int (cosudu/ cosu)
= -9 Int ( du/u) = -9 lnu = -9 ln [ arcsin ( x/3) ] + C

Answer 5

Thanks you for detect the typo --> huge mistake :")

Answer 6

Umm.....sqrt(x^2-9) does not equal -sqrt(9-x^2).....-1 can't be factored out of a square root. Otherwise you would get an imaginary number. Either way, right approach :)

Answer 7

thank you! could you please explain your steps? i'm a little confused when the u sub is made. also they asked me to integrate it from 3sqrt2 to 6...when the substitutions are made does this change the bounds?

Answer 8

yes. Although I personally do the indefinite integral first and then resbustitute x back in so as to not get confused.

Answer 9

okay! for my answer i got int (cosu)/(cosu) du ...i do not think this is the correct answer though. after i substitute cos^2(x) in the denom, that is the answer that i got. the coefficients that i took out of the integral was 3/3

Answer 10

Sorry, my PC and connection in an out so much!
Any way, the key is if the
Integral is 1/sqrt(9 - x^2), then let x = asinu is correct!

Answer 11

However in this case, the integrad switch the sign:
1/sqrt(x^2-9)

Answer 12

That's why I take out negative sign at first to switch it back!

Answer 13

But after I check the internet, they let x = sec u,
then everything just like I work out, expcept with secu.

Answer 14

Are you following?

Answer 15

yes, thank you. because i double checked, and i ended up getting du/sqrt(-cosu) which is not possible. i will try it with secu instead

Answer 16

I'll update with you!

Answer 17

question: would it be easier to integrate it by parts instead of trig sub?

Answer 18

No, that's the best method !

Answer 19

I'll just redo it!

Answer 20

Let x = 3secu --> dx = 3secu tanu du, x^2 = 9 sec^2u
-> sqrt ( x^2 - 9) = sqrt ( 9 sec^2u - 9)
= 3 sqrt ( sec^2u -1) = 3 tanu

Answer 21

Do you understand up to this point?

Answer 22

yes!

Answer 23

integral of dx/sqrt(x^2-9)
= Int ( 3 sec u tan u du/ 3 tanu )
= Int ( sec u ) du

Answer 24

ln [ tan u + secu ] + c = ln [ tan ( sec^1 (x/3) + x/3) ] + C

Answer 25

Do you agree with this result?

Answer 26

yes, however i don't understand how to find the last step. why is int 1/(secu) = ln [tanu+secu]

Answer 27

sec u =sec u(tan u+ sec u)/(tan u +sec u) = (sec u tan u +sec^2 u)/(tan u +sec u). The numerator is the derivative of the denominator, so using u substitution gets the desured integral.

Answer 28

oh! okay thank you!