# What is inttan(x)^3sec(x)^3dx ?

Aug 7, 2018

$\int \tan {\left(x\right)}^{3} \sec {\left(x\right)}^{3} \mathrm{dx} = \frac{1}{5} \sec {\left(x\right)}^{5} - \frac{1}{3} \sec {\left(x\right)}^{3} + C$

#### Explanation:

$I = \int \tan {\left(x\right)}^{3} \sec {\left(x\right)}^{3} \mathrm{dx}$

$= \int \tan \left(x\right) \tan {\left(x\right)}^{2} \sec {\left(x\right)}^{3} \mathrm{dx}$

Because $\tan {\left(x\right)}^{2} = \sec {\left(x\right)}^{2} - 1$

$I = \int \tan \left(x\right) \sec {\left(x\right)}^{3} \left(\sec {\left(x\right)}^{2} - 1\right) \mathrm{dx}$

$= \int \tan \left(x\right) \sec {\left(x\right)}^{5} \mathrm{dx} - \int \tan \left(x\right) \sec {\left(x\right)}^{3} \mathrm{dx}$

$= - \int \frac{- \sin \left(x\right)}{\cos {\left(x\right)}^{6}} \mathrm{dx} + \int \frac{- \sin \left(x\right)}{\cos {\left(x\right)}^{4}} \mathrm{dx}$

Let $u = \cos \left(x\right)$
$\mathrm{du} = - \sin \left(x\right) \mathrm{dx}$

So:

$I = \int \frac{1}{u} ^ 4 \mathrm{du} - \int \frac{1}{u} ^ 6 \mathrm{du}$

$= \frac{1}{5 {u}^{5}} - \frac{1}{3 {u}^{3}} + C$, $C \in \mathbb{R}$

$= \frac{1}{5 \cos {\left(x\right)}^{5}} - \frac{1}{3 \cos {\left(x\right)}^{3}} + C$, $C \in \mathbb{R}$

$= \frac{1}{5} \sec {\left(x\right)}^{5} - \frac{1}{3} \sec {\left(x\right)}^{3} + C$, $C \in \mathbb{R}$