How do you integrate #int e^-xtan(e^-x)dx#?

Answer 1

#inte^-xtan(e^-x)dx=ln(abscos(e^-x))+C#

#inte^-xtan(e^-x)dx#
First, let #u=e^-x#. Differentiating this shows that #du=-e^-xdx#.
#=-inttan(e^-x)(-e^-x)dx#
#=-inttan(u)du#

This is a standard integral, but we can show how to integrate it by rewriting tangent as sine divided by cosine:

#=-intsin(u)/cos(u)du#
Now, let #v=cos(u)#. This implies that #dv=-sin(u)du#.
#=int(-sin(u))/cos(u)du#
#=int(dv)/v#

This is also a standard (and very important) integral:

#=ln(absv)+C#
#=ln(abscos(u))+C#
#=ln(abscos(e^-x))+C#
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Answer 2

To integrate ( \int e^{-x} \tan(e^{-x}) , dx ), you can use substitution. Let ( u = e^{-x} ). Then, ( du = -e^{-x} , dx ), and solving for ( dx ), we get ( dx = -\frac{1}{u} , du ). Now, we rewrite the integral in terms of ( u ):

[ \int e^{-x} \tan(e^{-x}) , dx = \int -\tan(u) , du ]

Now, integrate ( -\tan(u) ) with respect to ( u ):

[ \int -\tan(u) , du = \ln|\sec(u)| + C ]

Substitute ( u = e^{-x} ) back in:

[ \ln|\sec(e^{-x})| + C ]

Therefore, the integral of ( \int e^{-x} \tan(e^{-x}) , dx ) is ( \ln|\sec(e^{-x})| + C ).

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Answer from HIX Tutor

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.

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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