# How do you find the indefinite integral of int tan5theta?

Nov 13, 2016

$\int \tan 5 \theta d \theta = \frac{1}{5} \ln | \sec 5 \theta | + C$

#### Explanation:

Let $u = 5 \theta \implies \frac{\mathrm{du}}{d \theta} = 5$
So, $\int \ldots \mathrm{du} = \int \ldots 5 d \theta$, or $\int \ldots d \theta = \int \ldots \frac{1}{5} \mathrm{du}$

Substituting into $\int \tan 5 \theta d \theta$ we get:

$\int \tan 5 \theta d \theta = \int \tan u \frac{1}{5} \mathrm{du}$
$\therefore \int \tan 5 \theta d \theta = \frac{1}{5} \int \tan u \mathrm{du}$
$\therefore \int \tan 5 \theta d \theta = \frac{1}{5} \ln | \sec u | + C$

And then replacing $u = 5 \theta$ gives:

$\int \tan 5 \theta d \theta = \frac{1}{5} \ln | \sec 5 \theta | + C$

Nov 13, 2016

$= - \frac{1}{5} \ln \cos 5 \theta + C$, in ${Q}_{1} \mathmr{and} {Q}_{3}$ and

$= - \frac{1}{5} \ln \left(- \cos 5 \theta\right)$, in ${Q}_{2} \mathmr{and} {Q}_{4}$..

#### Explanation:

$\tan 5 \theta$ is continuous for any interval sans (not including )

$5 \theta$=an odd multiple of $\frac{\pi}{2}$.

So, in any such interval.

the indefinite $\int \tan 5 \theta d \theta$

$= \int d \left(- \frac{1}{5} \ln \cos 5 \theta\right)$, in ${Q}_{1} \mathmr{and} {Q}_{3}$

$= - \frac{1}{5} \ln \cos 5 \theta + C$, in ${Q}_{1} \mathmr{and} {Q}_{3}$ and

=int d(-1/5ln(- cos 5theta), in ${Q}_{2} \mathmr{and} {Q}_{4}$

$= - \frac{1}{5} \ln \left(- \cos 5 \theta\right)$, in ${Q}_{2} \mathmr{and} {Q}_{4}$..

Note that as $\theta \to {\left(\left(2 k - 1\right) \frac{\pi}{10}\right)}_{\pm} ,$

$5 \theta \to {\left(\left(2 k - 1\right) \frac{\pi}{2}\right)}_{\pm} \mathmr{and} \tan 5 \theta \to \pm \infty$