## dy/dx = 1/(x.sqrt(x^2-1))

Hi,

Can anyone help me here!

dy/dx = 1/(x.sqrt(x^2-1)) , x>0 Note y = pi when x = 2

Here is what I did:

Let u = x^2 - 1
x^2 = u + 1

Substituting this in the above yields:

dy/dx = 1/x.sqrt(u)

Differentiating: u = x^2-1

du = 2x.dx

x = du/2.dx

Then,

dy/dx= 2/sqrt(u).du

I don't know if this is rite.
Plz help me!
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 Blog Entries: 9 Recognitions: Homework Help Science Advisor What are u trying to do?Find the antiderivative of $\frac{1}{x\sqrt{x^{2}-1}} [/tex] ...?If so,then use a substitution involving [itex] \cosh u$... Daniel.
 Why is it that everyone has been taught about hyperbolic trig functions when we skipped it in high school and calc 1-3?

## dy/dx = 1/(x.sqrt(x^2-1))

I was trying to simplify the Right Hand side, before, I could integrate , by making a substitution.

Anyway what would be the hyperbolic sub, for this involving coshu.

I don't have my text book with me. I am at work, and I am trying to finish my homework.
 Blog Entries: 9 Recognitions: Homework Help Science Advisor Need i say that your initial approach is incorrect...? It also works with $\sec u$... Daniel.

Blog Entries: 9
Recognitions:
Homework Help
 Blog Entries: 9 Recognitions: Homework Help Science Advisor $$\int \frac{\sec u \ \tan u \ du}{\sec u \ \tan u} =\int du = u+C = \mbox{arcsec} \ x +C$$ Daniel.
 Blog Entries: 9 Recognitions: Homework Help Science Advisor "sec" (as "cosine") is multivalued.But your could take $$\sec^{-1} 2=\frac{\pi}{3}$$ Daniel.