Integration using Hyperbolic Trig substitution

In summary, the integral \int\\{1}/{\sqrt{x^2-1}} dx between -3, -2 can be evaluated using hyperbolic substitution. By substituting x = -sinht and dx = cosht dt, the integral simplifies to int dt, which can then be integrated to get t + c. By substituting back for sinh(t) and using the trig identity cosh^2(t) - sinh^2(t) = 1, the final solution is given as (1 / (-sinh(-3))) - (1 / (-sinh(-2))) = -0.175898995.
  • #1
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Homework Statement



Evaluate:

[tex]\int\\{1}/{\sqrt{x^2-1}}[/tex] dx between -3, -2

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

Homework Equations



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

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 into 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:
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  • #2
your identity is incorrect, it's supposed to be cosh2t-sinh2t=1
 
  • #3
cheers.

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

then everything cancels and I am 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
this is the current solution I am 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:
  • #5
so if you have [itex]\int dt =t +c[/itex]

and x=cosh(t), doesn't it mean that t=cosh-1(x) ?
 
  • #6
Sorry, x = -sinht, because the domain of the integral is negative.

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

What is hyperbolic trig substitution?

Hyperbolic trig substitution is a method used in integration to simplify and solve integrals involving expressions with square roots of quadratic functions.

When is hyperbolic trig substitution used?

Hyperbolic trig substitution is typically used when dealing with integrals that involve expressions with square roots of quadratic functions, or when the integrand contains both a square root and a quadratic expression.

How does hyperbolic trig substitution work?

Hyperbolic trig substitution involves replacing the variable in the integral with a hyperbolic trigonometric function such as sinh, cosh, or tanh. This allows for the integration to be simplified and solved using known trigonometric identities.

What are the common hyperbolic trig substitutions?

The most commonly used hyperbolic trig substitutions are:
- x = sinh(u)
- x = cosh(u)
- x = tanh(u)

What are some tips for using hyperbolic trig substitution?

Some tips for using hyperbolic trig substitution include:
- Identifying which hyperbolic trigonometric function to use based on the form of the integral
- Simplifying the integral using trigonometric identities before applying hyperbolic trig substitution
- Making sure to properly substitute and substitute back in the original variable after solving the integral.

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