Question

\(\LARGE \int \frac{\sqrt{x^2 - 4}}{x} dx\)

Note: Trigonometric Substitution

I think i got as far as \(\LARGE 2\int \cot ^2 \theta d\theta\) though i do not know if it is on the right track

Answer 1

Ahaa. Hmm.
here, you're basically staring at your answer.

\[cot^2\theta = cosec^2\theta - 1\]


so your integral becomes "
\[\int cosec^2\theta.d\theta - \int1.d\theta\]


The derivative of -coxecx anyone?

Answer 2

ofcourse the '2' prefixed. and that will read '-cosec x', not "-coxecx"

Answer 3

that's it? o.O well is cot^2 on the right track?

Answer 4

Oh that I don't know, I was assuming you did it right till there, let me check.

Answer 5

put x=2sinu
dx=2cosu du
int (sqrt(x^2-4)/x)dx
=int(sqrt(4sin^2 u -4) /2sinu) (2cosu)du
=int(2cosu/2sinu)(2cosu)du
=2int (cotu)(cosu)du
:
:
:

Answer 6

it came from a \(\LARGE \int \frac{tan \theta}{\sec\theta} \sec \theta \tan \theta d \theta\)

Answer 7

Niet.

A wee wee bit of trouble. You did the substitution alright, but what did you do to 'dx' Did you make appropriate considerations for that?
What substitution did you make?

Answer 8

i used \(\LARGE x = 2\sec \theta\)

Answer 9

Very well. so ...(am writing 'A' instead of 'theta' that's easier)
dx = 2.secA.tanA.dA

so, did you replace that instead of dx?

Answer 10

yeah..as written in the above :P

Answer 11

so...am i right with cot ^2 x?

Answer 12

@apoorvk

Answer 13

of course by x i meant theta

Answer 14

Oh okay you're doing it right. I'm drunk.

Answer 15

ah i think i got it...

Answer 16

im ready for your cutsies now =_=