# Please give any hints about this integral

Hi guys, I and my colleague have been struggling with this integral for over 2 hours. :D

Any hints about solving it are welcome.

$$\int \frac{\sqrt{x^2 - 4}}{x^4}$$

Thanks in advance :)

## Answers and Replies

Dick
Science Advisor
Homework Helper
Looks like a trig substitution to me. Try u=2*sec(x).

Hi Dick,

thanks for your reply.

Unfortunately, in some countries(like mine) the secant and cosecant trig functions are not taught(not even in mathematical schools).

Aren't there any alternative approaches? Isn't there a way to use sin/cos/tg/cotg substitutions?

Thanks!

You could try $$u=2*sin^{-1}(x)$$

arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
The simplest is to use the hyperbolic cosine substitution, x=2Cosh(u).

Using this, you'll readily find a proper anti-derivative, namely:
$$A(x)=\frac{1}{12}(\frac{\sqrt{x^{2}-4}}{x})^{3}$$

Mark44
Mentor
Unfortunately, in some countries(like mine) the secant and cosecant trig functions are not taught(not even in mathematical schools).
Thanks!
I find it difficult to believe that these functions aren't presented in some countries. If that's the case, though, here is how they're defined:
secant(x) = sec(x) = 1/cos(x)
cosecant(x) = csc(x) = 1/sin(x)

Mark44
Mentor
You could try $$u=2*sin^{-1}(x)$$
Dagda, did you mean this as u = 2/sin(x)? That's different from 2*sin^(-1)(x), which is the same as 2*arcsin(x).

Dagda, did you mean this as u = 2/sin(x)? That's different from 2*sin^(-1)(x), which is the same as 2*arcsin(x).

I think my brain exploded and I accidentally read the wrong trig sub off the table, whatever happened you are of course correct. 