Integration using Hyperbolic Trig substitution

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Homework Help Overview

The discussion revolves around evaluating the integral \(\int \frac{1}{\sqrt{x^2-1}} \, dx\) using hyperbolic trigonometric substitution, specifically within the bounds of -3 and -2. Participants are exploring the implications of using hyperbolic functions in the context of the integral's negative bounds.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of hyperbolic substitution, with one suggesting \(x = -\cosh(t)\) and another considering \(x = -\sinh(t)\). There is uncertainty regarding the correct identities and substitutions to use, particularly in relation to the negative bounds of the integral.

Discussion Status

The conversation includes attempts to clarify the correct hyperbolic identities and substitutions. Some participants express uncertainty about the accuracy of their approaches and the implications of their substitutions. There is no explicit consensus on the final approach, but various lines of reasoning are being explored.

Contextual Notes

Participants note the constraints imposed by the negative bounds of the integral, which influence their choice of substitutions. There is also a correction regarding the hyperbolic identity used, indicating a need for careful consideration of the mathematical relationships involved.

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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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your identity is incorrect, it's supposed to be cosh2t-sinh2t=1
 
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
 
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:
so if you have [itex]\int dt =t +c[/itex]

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

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

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