Integration using cosh(t) or sinh(t)

  • Thread starter Thread starter Rubik
  • Start date Start date
  • Tags Tags
    Integration
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
5 replies · 4K views
Rubik
Messages
95
Reaction score
0
How do I go about solving the equation [tex]\int[/tex]dx/x squ rt of x^2 -1 using the appropriate substitution?
 
Physics news on Phys.org
first find a trig or hyper-trig identity that will simplify the intergand

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

then
[tex]x = sin(x)[/tex]
[tex]x = cos(x)[/tex]

are both good choices as
[tex]sin^2(x) = 1 - cos^2(x)[/tex]
[tex]cos^2(x) = 1 - sin^2(x)[/tex]

in your case you will need find something that simplifies
[tex]sqrt{1-x^2}[/tex]
 
Having [tex]\sqrt{x^2-1}[/tex] suggests a couple of possible substitutions:

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

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

Try either or both to see what works.
 
Rubik said:
How do I go about solving the equation [tex]\int[/tex]dx/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:

[tex]\int \frac{dx}{x \sqrt{x^2-1}}[/tex]

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