How do you integrate $\int e^{- x} tan (e^{- x}) d x$ $int e^-xtan(e^-x)dx$ ?

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Answer 1 · 2016-12-13T23:08:38

$\int e^{- x} tan (e^{- x}) d x = ln (∣ ∣ cos (e^{- x}) ∣ ∣) + C$ $inte^-xtan(e^-x)dx=ln(abscos(e^-x))+C$

$\int e^{- x} tan (e^{- x}) d x$ $inte^-xtan(e^-x)dx$

First, let $u = e^{- x}$ $u=e^-x$ . Differentiating this shows that $d u = - e^{- x} d x$ $du=-e^-xdx$ .

$= - \int tan (e^{- x}) (- e^{- x}) d x$ $=-inttan(e^-x)(-e^-x)dx$

$= - \int tan (u) d u$ $=-inttan(u)du$

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

$= - \int \frac{sin (u)}{cos (u)} d u$ $=-intsin(u)/cos(u)du$

Now, let $v = cos (u)$ $v=cos(u)$ . This implies that $d v = - sin (u) d u$ $dv=-sin(u)du$ .

$= \int \frac{- sin (u)}{cos (u)} d u$ $=int(-sin(u))/cos(u)du$

$= \int \frac{d v}{v}$ $=int(dv)/v$

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

$= ln (| v |) + C$ $=ln(absv)+C$

$= ln (| cos (u) |) + C$ $=ln(abscos(u))+C$

$= ln (∣ ∣ cos (e^{- x}) ∣ ∣) + C$ $=ln(abscos(e^-x))+C$

How do you integrate int e^-xtan(e^-x)dx∫e−xtan(e−x)dxint e^-xtan(e^-x)dx?