Question #9e12d

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Question #9e12d

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Answer 1 · 2017-03-22T19:52:36

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

Explanation:

I'll be using the common notation $tg (x) = tan (x)$ $"tg"(x)=tan(x)$ . So, you're asking for

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

Let $u = - e^{- x}$ $u=-e^-x$ . Differentiating this shows that $d u = e^{- x} d x$ $du=e^-xdx$ :

$= \int tan (- e^{- x}) (e^{- x} d x) = \int tan (u) d u$ $=inttan(-e^-x)(e^-xdx)=inttan(u)du$

Rewriting the tangent function:

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

Now let $v = cos (u)$ $v=cos(u)$ . This implies that $d v = - sin (u) d u$ $dv=-sin(u)du$ . Our integrand has both $v$ $v$ and is is a factor of $- 1$ $-1$ away from also containing $d v$ $dv$ :

$= - \int \frac{- sin (u)}{cos (u)} d u = - \int \frac{1}{v} d v$ $=-int(-sin(u))/cos(u)du=-int1/vdv$

This is a well-known and important antiderivative:

$= - ln | v |$ $=-lnabsv$

Returning to our original variable $x$ $x$ using $v = cos (u)$ $v=cos(u)$ and $u = - e^{- x}$ $u=-e^-x$ :

$= - ln | cos (u) | = - ln ∣ ∣ cos (- e^{- x}) ∣ ∣ + C$ $=-lnabscos(u)=-lnabscos(-e^-x)+C$

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Question #9e12d

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