Please give any hints about this integral

[bLaf]

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 :)

 

[Dick]

Looks like a trig substitution to me. Try u=2*sec(x).

 

[bLaf]

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!

 

[The Dagda]

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

 

[arildno]

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]

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]

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).

 

[The Dagda]

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.