Question

Integrate using trigonometric substitution:
x
------- dx?
(x^2 -4)

Answer 1

Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

Answer 2

lat x=2sectheta
\[\int\limits_{}^{}\frac{\sec \theta }{\tan \theta}\sec \theta \tan \theta d \theta\]

Answer 3

\[\int\limits_{}^{}\sec ^{2}\theta d \theta =\tan \theta +c \]

Answer 4

so the integral is what?

Answer 5

\[\int\limits_{}^{}(\sec \theta)\sec \theta \tan \theta d \theta /\tan^2 \theta =\int\limits_{}^{}\sec^2 \theta d \theta /\tan \theta\]

Answer 6

hold on is this just
\[\int \frac{x}{x^2-4}dx\]

Answer 7

Yes

Answer 8

and if so why is this a trig sub?

Answer 9

is it log(x^2-4)/2??

Answer 10

looks like a fairly simple partial fraction problem. i think you get \[\int\frac{1}{x-2}dx + \int\frac{1}{x+2}dx\]

Answer 11

oops no sorry there should be a 1/2 out front

Answer 12

one half of that whole thing.

Answer 13

Because that was the question I had in the list of practice questions

Answer 14

oh ok. well after you make the trig sub you are going to have to switch back

Answer 15

wait wait you do not even need partial fractions. just a simple u-sub

Answer 16

\[u=x^2-4\]
\[du=2xdx\] and you aer done in one step

Answer 17

answer is
\[\frac{1}{2}\ln(x^2-4)\]

Answer 18

maybe the teacher meant u - sub instead of trig sub

Answer 19

shine had it earlier, sorry

Answer 20

Does seem easier, thanks for the help :D

Answer 21

sorry about the false start. partial fractions indeed!