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

  • Thread starter bolas
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CIn summary, the conversation discusses how to rewrite the integral \frac{1}{\cos(x)-1}dx using trigonometric identities and substitution. The final answer is \int \frac{dx}{\cos x-1}=-\int \frac{dx}{1-\cos x}=-\frac{1}{2}\int \frac{dx}{\sin^{2}\frac{x}{2}} using the substitution T={\sqrt\frac{\a}{g}}\In\fract{a+\sqrt{a^2-b^2}}{b}+C.
  • #1
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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  • #2
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]
 
  • #3
[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.
 
  • #4
[tex]\\T=[\sqrt\frac{\a}{g}]\In\fract{a+\sqrt{a^2-b^2}}{b}[/tex]
 
Last edited:
  • #5
[tex]\\T={\sqrt\frac{\a}{g}}\In\fract{a+\sqrt{a^2-b^2}}{b}[/tex]
 

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

What is the purpose of solving $\frac{1}{\cos(x)-1}dx$?

The purpose of solving $\frac{1}{\cos(x)-1}dx$ is to find the indefinite integral of the given function. This will allow us to calculate the area under the curve of the function and use it in various applications such as physics, engineering, and economics.

What is the difficulty level of solving $\frac{1}{\cos(x)-1}dx$?

Solving $\frac{1}{\cos(x)-1}dx$ can be considered a moderately difficult integration problem. It requires knowledge of basic integration techniques, such as substitution and integration by parts, as well as familiarity with trigonometric identities.

Can you provide any hints for solving $\frac{1}{\cos(x)-1}dx$?

One helpful hint for solving $\frac{1}{\cos(x)-1}dx$ is to use the substitution $u = \tan(\frac{x}{2})$. This will transform the integral into a more manageable form. Additionally, using trigonometric identities, such as $\cos^2(x) = \frac{1}{2}(1+\cos(2x))$, can also be helpful.

Are there any common mistakes to avoid when solving $\frac{1}{\cos(x)-1}dx$?

One common mistake to avoid when solving $\frac{1}{\cos(x)-1}dx$ is forgetting to include the constant of integration. Another common mistake is not simplifying the final answer, which can lead to incorrect results.

Can you provide an example of solving $\frac{1}{\cos(x)-1}dx$?

Example: Find the indefinite integral of $\frac{1}{\cos(x)-1}dx$.
Solution: Using the substitution $u = \tan(\frac{x}{2})$, we can rewrite the integral as $\int\frac{1}{\frac{1-u^2}{1+u^2}}\cdot\frac{2}{1+u^2}du$. Simplifying, we get $\int\frac{2}{1-u^2}du$. Using partial fractions, we can rewrite this as $\int\frac{1}{1-u}du + \int\frac{1}{1+u}du$. Solving these integrals and substituting back in $x$, we get the final answer of $-\ln|\cos(x)-1| + \ln|\cos(x)+1|+C$.

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