# Thread: Integration Using Hyperbolic Identities

1. ## Integration Using Hyperbolic Identities

So i need to solve:

The integral of 1/ Sqrt( 9x^2 - 16) dx

If the 9 wasnt infront of x^2, i could easily do it, letting x = 4cosh(u) and so on, but the 9 confuses me.

Am i right in thinking i take a 9 out, to leave :

1/ sqrt( 9(x^2 - 16/9) ) dx and then: (1/3) * Integral of 1/ sqrt(x^2 - 16/9) dx?

As when i tried this, i got the incorrect answer.

Also, apologies if the format is difficult to understand, i tried to use latex, but got ' Latex Error: Compile failed'

Thanks

2. Originally Posted by Mcoolta
So i need to solve:

The integral of 1/ Sqrt( 9x^2 - 16) dx

If the 9 wasnt infront of x^2, i could easily do it, letting x = 4cosh(u) and so on, but the 9 confuses me.

Am i right in thinking i take a 9 out, to leave :

1/ sqrt( 9(x^2 - 16/9) ) dx and then: (1/3) * Integral of 1/ sqrt(x^2 - 16/9) dx?
So far so good. Now make the substitution x = sqrt(16/9)*cosh(u) = (4/3)*cosh(u). This gives dx = (4/3)*sinh(u) du, so the integral becomes
$\int \frac{dx}{\sqrt{9x^2 - 16}} = \frac{1}{3} \int \frac{(4/3)~sinh(u)~du}{\frac{4}{3} \sqrt{cosh^2(u) - 1}}$

$= \frac{1}{3} \int \frac{sinh(u)~du}{\sqrt{cosh^2(u) - 1}}$

Can you take it from here?

-Dan

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