# How do you find the antiderivative of #dx/(cos(x) - 1)#?

Do some conjugate multiplication, apply some trig, and finish to get a result of

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To find the antiderivative of ( \frac{dx}{\cos(x) - 1} ), you can use a trigonometric substitution. Let ( u = \tan\left(\frac{x}{2}\right) ). Then, ( dx = \frac{2}{1 + u^2} du ).

Substitute these into the integral: [ \int \frac{dx}{\cos(x) - 1} = \int \frac{2}{\frac{1 - u^2}{1 + u^2}} du ]

Simplify the expression inside the integral: [ = \int \frac{2(1 + u^2)}{1 - u^2} du ]

Now, perform partial fraction decomposition: [ \frac{2(1 + u^2)}{1 - u^2} = \frac{A}{1 + u} + \frac{B}{1 - u} ]

Solve for ( A ) and ( B ), then integrate each term separately to find the antiderivative. After integrating, you will obtain the final antiderivative expression in terms of ( x ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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