# How do you integrate #(1-tanx)/(1+tanx) dx#?

Rewrite as sines and cosines.

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To integrate ( \frac{{1 - \tan(x)}}{{1 + \tan(x)}} , dx ), you can use the substitution method. Let ( u = \tan(x) ), then ( du = \sec^2(x) , dx ). Substitute ( u ) and ( du ) into the integral and simplify. You'll get ( \int \frac{{1 - u}}{{1 + u}} \frac{{du}}{{\sec^2(x)}} ). Simplify this expression and integrate with respect to ( u ). After integrating, replace ( u ) with ( \tan(x) ) to obtain the final result.

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To integrate (\frac{1 - \tan(x)}{1 + \tan(x)} , dx), use the substitution method:

Let (u = \tan(x) + 1), then (du = \sec^2(x) , dx).

Substitute (u = \tan(x) + 1) and (du = \sec^2(x) , dx) into the integral:

[\int \frac{1 - \tan(x)}{1 + \tan(x)} , dx = \int \frac{1 - \tan(x)}{u} \cdot \frac{du}{\sec^2(x)}]

Now, simplify the expression:

[\int \frac{1 - \tan(x)}{1 + \tan(x)} , dx = \int \frac{1 - \tan(x)}{u} \cdot \cos^2(x) , du]

Next, integrate with respect to (u):

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

Now, integrate ( \frac{1 - u}{u} ) with respect to (u):

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

[= \ln|u| - u + C]

Finally, substitute back (u = \tan(x) + 1):

[= \ln|\tan(x) + 1| - (\tan(x) + 1) + C]

So, the integral of ( \frac{1 - \tan(x)}{1 + \tan(x)} , dx ) is ( \ln|\tan(x) + 1| - (\tan(x) + 1) + 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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