How Do You Integrate 1/(cos(x)-1)?

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SUMMARY

The integral of 1/(cos(x)-1) can be approached by multiplying the integrand by (cos(x)+1)/(cos(x)+1), resulting in the expression (cos(x)+1)/(-sin(x))^2. This leads to two separate integrals: the integral of cos(x)/(-sin(x))^2 and the integral of 1/(-sin(x))^2. The latter simplifies to -cot(x), while the former can be expressed as -∫(cot(x)csc(x))dx. The discussion emphasizes the importance of verifying antiderivatives by differentiation.

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Homework Statement



help me out. I want to find the integral of 1/(cosx-1). I first multiplied the function by cosx+1/cosx+1. then i ended up with the integral of (cosx+1)/(-sinx)^2 leading to integral of cosx/(-sinx)^2 plus integral of 1/(-sinx)^2.

the integral of 1/(-sinx)^2 i believe is just cotx.

the integral of cosx/(-sinx)^2 i think is -(integral of cotx*cscx)


Homework Equations



integral of 1/(cosx-1)
 
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What's your question? I don't see anything wrong in what you've described. You can check any antiderivative you get by taking its derivative. If you get back to the integrand, you're good.
 
What function has as its derivative -cot x csc x?
 

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