Question

integrate 1/ (t^3 (Square root (t^2-4) (dt) with a range of 4, (2 square root of 2). These ranges are next to the integrate sign. If anyone one could help me this would be great. Thanks.

Answer 1

trig sub for this one

Answer 2

it would be the sec trig sub correct. And if so my question is with the numbers well ranges that are given how do i work this into the problem.

Answer 3

change the limits of integration when you make the sub
then you don't have to change back

Answer 4

Okay. I did that and got 4 to change into Pie over 3 but i am confused about the 2 square root of 2 how this one changes. I feel i have it wrong.

Answer 5

sub is \(x=\sec(\theta)\) i think so \(\theta=\sec^{-1}(x)\)

Answer 6

Oh okay so would that mean my pie over 3 is incorrect?

Answer 7

\[\int _4^{2\sqrt{2}}\frac{dt}{t^3\sqrt{t^2-1}}\] is that it?

Answer 8

Yes but the ranges are swapped.

Answer 9

oh and the 1 is a 4 that is in the square root under the denominator

Answer 10

or maybe this \[\int _{2\sqrt{2}}^4\frac{dt}{t^3\sqrt{t^2-1}}\]

Answer 11

oooh ok that is better \[\int _{2\sqrt{2}}^4\frac{dt}{t^3\sqrt{t^2-4}}\]

Answer 12

yes this would be my problem

Answer 13

that makes the sub
\[x=2\sec(\theta)\] and so \(\theta=\sec^{-1}(\frac{x}{2})\)

Answer 14

that makes a lot more sense
now you can compute the inverse easier

Answer 15

thank you. This should now help me figure out where i have gone wrong within my problem. So thank you.

Answer 16

lower limit is easy right? i get \(\frac{\pi}{4}\)

Answer 17

yw

Answer 18

yes i got \[\pi/4\] for the bottom

Answer 19

k good
top is probably easy too
my guess \(\frac{\pi}{3}\) without even looking as the problem as been cooked up to be easy (sort of)

Answer 20

Ya i got that too. But i am stuck at the ending if what i got is on the right track i dont know how to take it further to get the answer which i was given on my homework site. I feel like i am missing something, just dont know what?

Answer 21

what did you get when you made the sub?

Answer 22

\[\frac{2\sec(\theta)\tan(\theta)}{16\sec^3(\theta)\tan(\theta)}\] i think is the first step

Answer 23

Yes i got this part

Answer 24

leaving
\[\frac{1}{8}\cos^2(\theta)\]

Answer 25

Yes this is where i am now confused on how to take this further.

Answer 26

oooh this is the easy gimmick

Answer 27

\[\cos^2(\theta)=\frac{1}{2}(1+\cos(2\theta))\]

Answer 28

and that is easy to integrate

Answer 29

Thanks so much!

Answer 30

yw

Answer 31

okay sorry to bother you again. Is there anyway you can walk me through the rest of the steps for the problem cause i thought i saw were i needed to take this problem but it is not working out for me. If you could help i would appreciate it.

Answer 32

Never mind i figured it out thanks though. And again thanks for all the help.