# How do you integrate int x^3 t an x dx  using integration by parts?

Jan 7, 2016

$\int {x}^{3} \tan \left(x\right) \mathrm{dx} = \ln | \cos \left(x\right) | \left({x}^{3} - 3 {x}^{2} + 6 x - 6 + c\right)$

#### Explanation:

Say $\mathrm{dv} = \tan \left(x\right)$ so $v = \ln | \cos \left(x\right) |$
And $u = {x}^{3}$ so $\mathrm{du} = 3 {x}^{2}$

$\int {x}^{3} \tan \left(x\right) \mathrm{dx} = {x}^{3} \cdot \ln | \cos \left(x\right) | - 3 \int \tan \left(x\right) \cdot {x}^{2} \mathrm{dx}$

For the latter integral, repeat

$\mathrm{dv} = \tan \left(x\right)$ so $v = \ln | \cos \left(x\right) |$, $u = {x}^{2}$ so $\mathrm{du} = 2 x$

$\int {x}^{3} \tan \left(x\right) \mathrm{dx} = {x}^{3} \cdot \ln | \cos \left(x\right) | - 3 \left({x}^{2} \ln | \cos \left(x\right) | - 2 \int \tan \left(x\right) x \mathrm{dx}\right)$
$\int {x}^{3} \tan \left(x\right) \mathrm{dx} = {x}^{3} \cdot \ln | \cos \left(x\right) | - 3 {x}^{2} \ln | \cos \left(x\right) | + 6 \int \tan \left(x\right) x \mathrm{dx}$

And once more to finish it

$\mathrm{dv} = \tan \left(x\right)$ so $v = \ln | \cos \left(x\right) |$, $u = x$ so $\mathrm{du} = 1$

$\int {x}^{3} \tan \left(x\right) \mathrm{dx} = {x}^{3} \ln | \cos \left(x\right) | - 3 {x}^{2} \ln | \cos \left(x\right) | + 6 \left(x \ln | \cos \left(x\right) | - \ln | \cos \left(x\right) | + c\right)$

$\int {x}^{3} \tan \left(x\right) \mathrm{dx} = {x}^{3} \ln | \cos \left(x\right) | - 3 {x}^{2} \ln | \cos \left(x\right) | + 6 x \ln | \cos \left(x\right) | - 6 \ln | \cos \left(x\right) | + c$

Or

$\int {x}^{3} \tan \left(x\right) \mathrm{dx} = \ln | \cos \left(x\right) | \left({x}^{3} - 3 {x}^{2} + 6 x - 6 + c\right)$