Integral help

  • Thread starter bolas
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  • #1
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Main Question or Discussion Point

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
 

Answers and Replies

  • #2
19
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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]
 
  • #3
dextercioby
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[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
21
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[tex]\\T=[\sqrt\frac{\a}{g}]\In\fract{a+\sqrt{a^2-b^2}}{b}[/tex]
 
Last edited:
  • #5
21
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[tex]\\T={\sqrt\frac{\a}{g}}\In\fract{a+\sqrt{a^2-b^2}}{b}[/tex]
 

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