Solving $\frac{1}{\cos(x)-1}dx$ - Any Hints?

  • Context: Undergrad 
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Discussion Overview

The discussion revolves around solving the integral of the expression $\frac{1}{\cos(x)-1}dx$. Participants explore various methods and transformations to simplify the integral, including trigonometric identities and substitutions.

Discussion Character

  • Mathematical reasoning
  • Exploratory
  • Homework-related

Main Points Raised

  • One participant suggests rewriting the integral using the identity $1 - \cos^2 = \sin^2$ and attempts to manipulate the expression by multiplying by $\frac{\cos x + 1}{\cos x + 1}$.
  • Another participant proposes that the integral can be expressed as $\int \frac{\cos x + 1}{-\sin^2 x} dx$ and recommends using trigonometric identities to further transform it into $\int - \cot x \csc x - \csc^2 x dx$.
  • A different participant rewrites the integral as $-\int \frac{dx}{1-\cos x}$ and suggests using a substitution to relate it to $\int \frac{dx}{\sin^{2}\frac{x}{2}}$.
  • Two posts appear to contain expressions related to a different topic, possibly indicating a misunderstanding or off-topic contributions.

Areas of Agreement / Disagreement

Participants present multiple approaches to the integral, indicating a lack of consensus on the best method to solve it. The discussion remains unresolved as different techniques are proposed without agreement on a final solution.

Contextual Notes

Some participants rely on specific trigonometric identities and substitutions, which may depend on the definitions and assumptions they are using. The discussion does not clarify all steps or assumptions involved in the transformations.

bolas
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I need help with this one:

[tex]\frac{1}{\cos(x)-1}dx[/tex]

I know there must be a way to rewrite it so that i can use 1 - cos^2 = sin^2. I've tried to multiply by cosx + 1 / cosx + 1 and that gets me further but the answer doesn't seem to be the same maybe because of simplification?

Any hints?

thanks
 
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The integral is equal to:

[itex]\int \frac {\cos x + 1}{-\sin ^2 x} dx[/itex] I suggest you use some trigonometric identities to change it into
[itex]\int - \cot x \csc x - \csc ^2 x dx[/itex]
 
[tex]\int \frac{dx}{\cos x-1}=-\int \frac{dx}{1-\cos x}=-\frac{1}{2}\int \frac{dx}{\sin^{2}\frac{x}{2}}[/tex]

use a simple substitution and a tabulated integral to get the answer.
 
[tex]\\T=[\sqrt\frac{\a}{g}]\In\fract{a+\sqrt{a^2-b^2}}{b}[/tex]
 
Last edited:
[tex]\\T={\sqrt\frac{\a}{g}}\In\fract{a+\sqrt{a^2-b^2}}{b}[/tex]
 

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