- #1

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Any hints about solving it are welcome.

[tex]\int \frac{\sqrt{x^2 - 4}}{x^4}[/tex]

Thanks in advance :)

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- Thread starter bLaf
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- #1

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Any hints about solving it are welcome.

[tex]\int \frac{\sqrt{x^2 - 4}}{x^4}[/tex]

Thanks in advance :)

- #2

Dick

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

- #3

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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!

- #4

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You could try [tex]u=2*sin^{-1}(x)[/tex]

- #5

arildno

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Using this, you'll readily find a proper anti-derivative, namely:

[tex]A(x)=\frac{1}{12}(\frac{\sqrt{x^{2}-4}}{x})^{3}[/tex]

- #6

Mark44

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

Thanks!

secant(x) = sec(x) = 1/cos(x)

cosecant(x) = csc(x) = 1/sin(x)

- #7

Mark44

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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).You could try [tex]u=2*sin^{-1}(x)[/tex]

- #8

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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.

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