nemired9

2021-12-18

What is the antiderivative of $\mathrm{tan}\left(x\right)$ ?

Papilys3q

Recall:
$\int \frac{{g}^{\prime }\left(x\right)}{g\left(x\right)}dx=\mathrm{ln}|g\left(x\right)|+C$
(You can verify this by substitution $u=g\left(x\right)$.)
Now, let us look at the posted antiderivative.
By the trig identity $\mathrm{tan}x=\frac{\mathrm{sin}x}{\mathrm{cos}x}$
$\int \mathrm{tan}xdx=\int \frac{\mathrm{sin}x}{\mathrm{cos}x}dx$
by rewriting it a bit further to fit the form above,
$=-\int \frac{-\mathrm{sin}x}{\mathrm{cos}x}dx$
by the formula above,
$=-\mathrm{ln}|\mathrm{cos}x|+C$
or by $r\mathrm{ln}x={\mathrm{ln}x}^{r}$
$=\mathrm{ln}{|\mathrm{cos}x|}^{-1}+C=\mathrm{ln}|\mathrm{sec}x|+C$
I hope that this was helpful.

Foreckije

By $\mathrm{tan}x=\frac{\mathrm{sin}x}{\mathrm{cos}x}$
$\int \mathrm{tan}xdx=\int \frac{\mathrm{sin}x}{\mathrm{cos}x}dx$
Let $u=\mathrm{cos}x⇒\frac{du}{dx}=-\mathrm{sin}x⇒dx=\frac{du}{-\mathrm{sin}x}$
By substitution,
$=\int \frac{\mathrm{sin}x}{u}\cdot \frac{du}{-\mathrm{sin}x}$
By cancelling $\mathrm{sin}x$
$=-\int \frac{1}{u}du$
By finding an antiderivative,
$=-\mathrm{ln}|u|+C$
By plugging $\mathrm{cos}x$ back in for u
$=-\mathrm{ln}|\mathrm{cos}x|+C$
By the log property $r\mathrm{ln}x={\mathrm{ln}x}^{r}$
$=\mathrm{ln}{|\mathrm{cos}x|}^{-1}+C$
By ${\left(\mathrm{cos}x\right)}^{-1}=\frac{1}{\mathrm{cos}x}=\mathrm{sec}x$
$=\mathrm{ln}|\mathrm{sec}x|+C$

RizerMix

We want to find $\int \mathrm{tan}xdx$
$\int \mathrm{tan}xdx=\int \frac{\mathrm{tan}x\mathrm{sec}x}{\mathrm{sec}x}dx$
Now let $u=\mathrm{sec}x$ and $du=\mathrm{sec}x\mathrm{tan}xdx$. Then
$\int \frac{\mathrm{tan}x\mathrm{sec}x}{\mathrm{sec}x}dx=\int \frac{1}{u}du$
This is a standard integral which evaluates to
$\mathrm{ln}|u|+C=\mathrm{ln}|\mathrm{sec}x|+C$

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