Integration using cosh(t) or sinh(t)

[Rubik]

How do I go about solving the equation \intdx/x squ rt of x^2 -1 using the appropriate substitution?

 

[lanedance]

first find a trig or hyper-trig identity that will simplify the intergand

for example if you had
\int dx \sqrt{1-x^2}

then
x = sin(x)
x = cos(x)

are both good choices as
sin^2(x) = 1 - cos^2(x)
cos^2(x) = 1 - sin^2(x)

in your case you will need find something that simplifies
sqrt{1-x^2}

 

[SammyS]

Having \sqrt{x^2-1} suggests a couple of possible substitutions:

\sec^2(\theta)-1=\tan^2(\theta),\ \ \text{so}\ \ x=\sec(t) makes sense for a trig substitution.

\cosh^2(u)-1=\sinh^2(u),\ \ \text{so}\ \ x=\cosh(t) makes sense for a hyperbolic function substitution.

Try either or both to see what works.

 

[Char. Limit]

How do I go about solving the equation \intdx/x squ rt of x^2 -1 using the appropriate substitution?

Just want to clarify, just what IS your integrand? It's very difficult to read. My best guess as to what it is is:

\int \frac{dx}{x \sqrt{x^2-1}}

Is that correct?

 

[Rubik]

Yes that is it :)

 

[Char. Limit]

In that case I would definitely use the substitution x=sec(u). You'll find that everything cancels out very nicely.