# Integrate x^3/sqrt((x^2+16)) dx using trig substitution?

Jan 28, 2018

$\int {x}^{3} / \sqrt{{x}^{2} + 16} \mathrm{dx} = \frac{1}{3} {\left(\frac{\sqrt{16 + {x}^{2}}}{x}\right)}^{3} - \frac{\sqrt{16 + {x}^{2}}}{x} + c$

#### Explanation:

Let $x = 4 \tan u$, then $\mathrm{dx} = 4 {\sec}^{2} u \mathrm{du}$

also $\sec u = \sqrt{1 + {x}^{2} / 16} = \frac{\sqrt{16 + {x}^{2}}}{x}$

and $I = \int {x}^{3} / \sqrt{{x}^{2} + 16} \mathrm{dx}$

= $\int \frac{64 {\tan}^{3} x}{\sqrt{16 {\tan}^{2} u + 16}} \times 4 {\sec}^{2} u \mathrm{du}$

= $\int \frac{64 {\tan}^{3} x}{4 \sec u} \times 4 {\sec}^{2} u \mathrm{du}$

= $\int 64 {\tan}^{3} u \sec u \mathrm{du}$

= $64 \int \left({\sec}^{2} u - 1\right) \left(\sec u \tan u\right) \mathrm{du}$

Let $v = \sec u$. Then $\mathrm{dv} = \sec u \tan u \mathrm{du}$:

and $I = \int \left({v}^{2} - 1\right) \mathrm{dv}$

= $\frac{1}{3} {v}^{3} - v + C$

= $\frac{1}{3} {\sec}^{3} u - \sec u + C$

= $\frac{1}{3} {\left(\frac{\sqrt{16 + {x}^{2}}}{x}\right)}^{3} - \frac{\sqrt{16 + {x}^{2}}}{x} + c$

Jan 28, 2018

$= \frac{4}{3} \sqrt{{x}^{2} + 4} \left({x}^{2} - 8\right)$

#### Explanation:

Try $x = 4 \tan \theta$. Then $x = 4 {\sec}^{2} \theta d \theta$
So
$\int {x}^{3} / \sqrt{{x}^{2} + 16} \mathrm{dx} = \int \frac{64 {\tan}^{3} \theta \times 4 {\sec}^{2} \theta}{4 \sec \theta} d \theta$
$= 64 \int {\tan}^{3} \theta \sec \theta d \theta = 64 \int {\tan}^{2} \theta \tan \theta \sec \theta d \theta$
$= 64 \int \left({\sec}^{2} \theta - 1\right) d \left(\sec \theta\right) = 64 \left(\frac{{\sec}^{3} \theta}{3} - \sec \theta\right)$
$= \frac{64}{3} \sec \theta \left({\sec}^{2} \theta - 3\right) = \frac{64}{3} \sqrt{1 + {x}^{2} / 16} \left(1 + {x}^{2} / 16 - 3\right)$
$= \frac{1}{3} \sqrt{{x}^{2} + 16} \left({x}^{2} - 32\right)$