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

Naeem

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!

dextercioby

What are u trying to do?Find the antiderivative of [itex] \frac{1}{x\sqrt{x^{2}-1}} [/tex] ...?If so,then use a substitution involving [itex] \cosh u [/itex]...

Daniel.

whozum

Why is it that everyone has been taught about hyperbolic trig functions when we skipped it in high school and calc 1-3?

Naeem

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.

dextercioby

Need i say that your initial approach is incorrect...?

It also works with [itex] \sec u [/itex]...

Daniel.

dextercioby

I wouldn't know.Self-taught hyperbolic trigonometry in HS (XII-th grade) when i had to evaluate antiderivatives using substitution...

Daniel.

Naeem

Ok, then,

Let x = secu

dx/du = secu.tanu

sec^2u-tan^2u = 1

dy/dx = 1/secu.tanu


Then what, to do!

dextercioby

[tex] \int \frac{\sec u \ \tan u \ du}{\sec u \ \tan u} =\int du = u+C = \mbox{arcsec} \ x +C [/tex]

Daniel.

Naeem

Ok, I applied the initial conditions:

y = pi when x = 2

and I got,

y = sec-1(x) + pi - sec-1(2)

can we write this as:

y = sec-1 (x/2) + pi

dextercioby

"sec" (as "cosine") is multivalued.But your could take

[tex] \sec^{-1} 2=\frac{\pi}{3} [/tex]

Daniel.