jamessinatraaa

2022-01-02

Evaluate the following integral.
$\frac{{e}^{\mathrm{tan}x}}{{\mathrm{cos}}^{2}x}dx$

movingsupplyw1

Expert

Step 1
Given integral is
$\frac{{e}^{\mathrm{tan}x}}{{\mathrm{cos}}^{2}x}dx$
We know that $\mathrm{sec}x=\frac{1}{\mathrm{cos}x}$
Therefore, given integral can be written as
$\int \frac{{e}^{\mathrm{tan}x}}{{\mathrm{cos}}^{2}x}dx=\int {e}^{\mathrm{tan}x}\left({\mathrm{sec}}^{2}xdx\right)$
Step 2
Now, we will use the following substitution to evaluate the integral
$\mathrm{tan}x=t$
${\mathrm{sec}}^{2}xdx=dt$
Therefore, we have
$\int \frac{{e}^{\mathrm{tan}x}}{{\mathrm{cos}}^{2}x}dx=\int {e}^{t}dt$
$\int \frac{{e}^{\mathrm{tan}x}}{{\mathrm{cos}}^{2}x}dx={e}^{t}+c$
$\int \frac{{e}^{\mathrm{tan}x}}{{\mathrm{cos}}^{2}x}dx={e}^{\mathrm{tan}x}+c$
Step 3
Ans:
$\int \frac{{e}^{\mathrm{tan}x}}{{\mathrm{cos}}^{2}x}dx={e}^{\mathrm{tan}x}+c$

Annie Gonzalez

Expert

Given:
$\int \frac{{e}^{\mathrm{tan}\left(x\right)}}{{\mathrm{cos}\left(x\right)}^{2}}dx$
$\int 1dt$
Use $\int 1dx=x$ to evaluate the integral
t
Substitute back
${e}^{\mathrm{tan}\left(x\right)}$
${e}^{\mathrm{tan}\left(x\right)}+C$

Vasquez

Expert

$\int \frac{{e}^{\mathrm{tan}\left(x\right)}}{\mathrm{cos}\left(x{\right)}^{2}}dx$
We put the expression $1/\mathrm{cos}\left(x{\right)}^{2}$ under the differential sign, i.e.:
$\frac{1}{\mathrm{cos}\left(x{\right)}^{2}}dx=d\left(\mathrm{tan}\left(x\right)\right),t=\mathrm{tan}\left(x\right)$
Then the original integral can be written as follows:
$\int {e}^{t}dt$
This is a tabular integral:
$\int {e}^{t}dt={e}^{t}+C$
To write down the final answer, it remains to substitute $\mathrm{tan}\left(x\right)$ instead of t.
${e}^{\mathrm{tan}\left(x\right)}+C$

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