# What's the integral of #int 1/(secx+tanx)dx#?

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The integral of ( \frac{1}{\sec(x) + \tan(x)} ) with respect to ( x ) can be solved by using a substitution method. Let ( u = \sec(x) + \tan(x) ). Then, ( du = (\sec(x)\tan(x) + \sec^2(x))dx ).

Substituting ( u ) and ( du ) into the integral, we have:

[ \int \frac{1}{\sec(x) + \tan(x)} , dx = \int \frac{1}{u} , du ]

This simplifies to:

[ \ln|u| + C ]

Substituting back ( u = \sec(x) + \tan(x) ), the final answer is:

[ \ln|\sec(x) + \tan(x)| + C ]

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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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