# Using trigonometric substitution, how do you integrate integral of x^3 (x^2+4)^(1/2) ?

Jul 21, 2018

$\int {x}^{3} {\left({x}^{2} + 4\right)}^{\frac{1}{2}} \mathrm{dx} = \frac{1}{5} {\left(4 + {x}^{2}\right)}^{2} \sqrt{4 + {x}^{2}} - \frac{4}{3} \left(4 + {x}^{2}\right) \sqrt{4 + {x}^{2}}$

See explanations below.

#### Explanation:

$\int {x}^{3} {\left({x}^{2} + 4\right)}^{\frac{1}{2}} \mathrm{dx} = \int {x}^{3} \sqrt{\left({x}^{2} + 4\right)} \mathrm{dx}$

let $x = 2 \tan \left(\theta\right)$

$\mathrm{dx} = 2 \sec {\left(\theta\right)}^{2} d \theta$

So :

$\int {x}^{3} {\left({x}^{2} + 4\right)}^{\frac{1}{2}} \mathrm{dx} = \int 16 \tan {\left(\theta\right)}^{3} \sqrt{\left(4 \tan {\left(\theta\right)}^{2} + 4\right)} \sec {\left(\theta\right)}^{2} d \theta$

$= 16 \int \tan {\left(\theta\right)}^{3} \sqrt{4 \left(\tan {\left(\theta\right)}^{2} + 1\right)} \sec {\left(\theta\right)}^{2} d \theta$

$= 32 \int \tan {\left(\theta\right)}^{3} {\cancel{\sqrt{\tan {\left(\theta\right)}^{2} + 1}}}^{\textcolor{red}{= \sec \left(\theta\right)}} \sec {\left(\theta\right)}^{2} d \theta$

$= 32 \int \tan {\left(\theta\right)}^{3} \sec {\left(\theta\right)}^{3} d \theta$

$= 32 \int \frac{\sin {\left(\theta\right)}^{3}}{\cos {\left(\theta\right)}^{6}} d \theta$

$= - 32 \int \frac{- \sin \left(\theta\right) \sin {\left(\theta\right)}^{2}}{\cos {\left(\theta\right)}^{6}} d \theta$

$= - 32 \int \frac{- \sin \left(\theta\right) \left(1 - \cos {\left(\theta\right)}^{2}\right)}{\cos {\left(\theta\right)}^{6}} d \theta$

let $u = \cos \left(\theta\right)$
$\mathrm{du} = - \sin \left(\theta\right) d \theta$

So : $32 \int \tan {\left(\theta\right)}^{3} \sec {\left(\theta\right)}^{3} d \theta = - 32 \int \frac{1 - {u}^{2}}{{u}^{6}} \mathrm{du}$

$= 32 \int {u}^{- 4} \mathrm{du} - 32 \int {u}^{- 6} \mathrm{du}$

$= \frac{32}{5} {u}^{- 5} - \frac{32}{3} {u}^{- 3}$

$= \frac{32}{5} \cos {\left(\theta\right)}^{- 5} - \frac{32}{3} \cos {\left(\theta\right)}^{-} 3$

Finally, because $\theta = \arctan \left(\frac{x}{2}\right)$ and $\cos \left(\arctan \left(\frac{x}{2}\right)\right) = \frac{2}{\sqrt{4 + {x}^{2}}}$,

$\int {x}^{3} {\left({x}^{2} + 4\right)}^{\frac{1}{2}} \mathrm{dx} = \frac{1}{5} {\left(4 + {x}^{2}\right)}^{2} \sqrt{4 + {x}^{2}} - \frac{4}{3} \left(4 + {x}^{2}\right) \sqrt{4 + {x}^{2}}$

\0/ Here's our answer !