# Evaluate the following integrals. \int e^{x}\tan(e^{x})dx

Evaluate the following integrals.
$\int {e}^{x}\mathrm{tan}\left({e}^{x}\right)dx$
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kalupunangh
Step 1
The given integral is $\int {e}^{x}\mathrm{tan}\left({e}^{x}\right)dx$
Step 2
Let $u={e}^{x}$.
Then $du={e}^{x}dx$
$\int {e}^{x}\mathrm{tan}\left({e}^{x}\right)dx=\int \mathrm{tan}udu$
$=\int \frac{\mathrm{sin}u}{\mathrm{cos}u}du$

$=-\mathrm{ln}|v|$
$=-\mathrm{ln}|\mathrm{cos}u|$
$=-\mathrm{ln}|\mathrm{cos}\left({e}^{x}\right)|+C$
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Cheryl King
$\int {e}^{x}{\mathrm{tan}e}^{x}dx$
Let $u={e}^{x}⇒du=-{e}^{x}⇒{e}^{x}dx=-du$
Apply the substitution
$\int {\mathrm{tan}e}^{x}{e}^{x}dx⇒-\int \mathrm{tan}udu$
Integrate, apply $\int \mathrm{tan}udu=-\mathrm{ln}|\mathrm{cos}u|+C$
$-\int \mathrm{tan}udu=-\left(-\mathrm{ln}|\mathrm{cos}u|\right)+C$
$=\mathrm{ln}|\mathrm{cos}u|+C$
Back - substitute $u={e}^{x}$
$=\mathrm{ln}\mid {\mathrm{cos}e}^{x}\mid +C$
Result:
$\mathrm{ln}\mid {\mathrm{cos}e}^{x}\mid +C$