# Integration using Hyperbolic Trig substitution

1. Mar 15, 2009

### Krash

1. The problem statement, all variables and given/known data

Evaluate:

$$\int\\{1}/{\sqrt{x^2-1}}$$ dx between -3, -2

I know I'm supposed to use hyperbolic substitution in the question.

2. Relevant equations

edit: cosh^2(t) - sinh^2(t) = 1

3. The attempt at a solution

let x = -cosht, inside the integral let dx = sinh(t) dt

int ( sinht / sqrt(cosh^2t - 1) ) dt

now normally at this point i would use the trig identity cosh^2(t) + sinh^2(t) = 1 to eliminate the -1 in the root, however i can't work out how to sub in to eliminate because i've had to use a negative cosht for x as the bounds of the integral are both negative.

I'm not great at this area of maths, I thought maybe if i substituted sinh(t) for x, I could possible rewrite -sinh(t) as sinh(-t) but i'm not sure if that's a fair solution or not.

Last edited: Mar 15, 2009
2. Mar 15, 2009

### rock.freak667

your identity is incorrect, it's supposed to be cosh2t-sinh2t=1

3. Mar 15, 2009

### Krash

cheers.

that still means i can't sub cosht for x but i can use sinh = x

then everything cancels and im left with int dt
which is -sinh^-1(x)

which gives me:

(1 / (-sinh(-3))) - (1 / (-sinh(-2))) = -0.175898995

still not sure if this is technically accurate

4. Mar 15, 2009

### Krash

this is the current solution im working on:

dx = cosht dt
x = -sinht

therefore;

1/ sqrt(x^2 - 1 ) = cosht dt / sqrt( -sinh^2 t - 1 ) ----- (1)
using the trig id:
cosh^2 t - sinh^2 t = 1
-sinh^2 t = -cosh^2 t + 1 ----- (2)

(2) into (1)

int (cosht / sqrt(-cosh^2 t)) dt = int dt
int dt = t + c
t = 1/(-sinh x)

therefore the integral evaluated is:
(1 / (-sinh(-3))) - (1 / (-sinh(-2))) = -0.175898995

Last edited: Mar 15, 2009
5. Mar 15, 2009

### rock.freak667

so if you have $\int dt =t +c$

and x=cosh(t), doesn't it mean that t=cosh-1(x) ?

6. Mar 15, 2009

### Krash

Sorry, x = -sinht, because the domain of the integral is negative.

i switched around the dx and x substitutions the second time around.