# How do you integrate int 1/(x^4sqrt(16+x^2)) by trigonometric substitution?

Mar 23, 2018

The answer is $= \frac{\sqrt{16 + {x}^{2}}}{256 x} - {\left({x}^{2} + 16\right)}^{\frac{3}{2}} / \left(768 {x}^{3}\right) + C$

#### Explanation:

Let $x = 4 \tan \theta$, $\implies$, $\mathrm{dx} = 4 {\sec}^{2} \theta d \theta$

$\sqrt{16 + {x}^{2}} = \sqrt{16 + 16 {\tan}^{2} \theta} = 4 \sec \theta$

Therefore, the integral is

$I = \int \frac{\mathrm{dx}}{{x}^{4} \sqrt{16 + {x}^{2}}} = \int \frac{4 {\sec}^{2} \theta d \theta}{256 {\tan}^{4} \theta \cdot 4 \sec \theta}$

$= \frac{1}{256} \int \frac{\sec \theta d \theta}{{\tan}^{4} \theta}$

$\tan \theta = \sin \frac{\theta}{\cos} \theta$

$\sec \theta = \frac{1}{\cos} \theta$

$I = \frac{1}{256} \int \frac{d \theta}{\cos \theta \cdot {\sin}^{4} \frac{\theta}{\cos} ^ 4 \theta}$

$= \frac{1}{256} \int \frac{{\cos}^{3} \theta d \theta}{{\sin}^{4} \theta}$

$= \frac{1}{256} \int \frac{\cos \theta \left(1 - {\sin}^{2} \theta\right) d \theta}{\sin} ^ 4 \theta$

Let $v = \sin \theta$, $\implies$, $\mathrm{dv} = \cos \theta d \theta$

$I = \frac{1}{256} \int \frac{\left(1 - {v}^{2}\right) \mathrm{dv}}{{v}^{4}}$

$= \frac{1}{256} \int \left(\frac{1}{v} ^ 4 - \frac{1}{v} ^ 2\right) \mathrm{dv}$

$= - \frac{1}{256} \cdot \frac{1}{3 {v}^{3}} + \frac{1}{256} \cdot \left(\frac{1}{v}\right)$

$= \frac{1}{256} \cdot \frac{1}{\sin} \theta - \frac{1}{768} \cdot \frac{1}{\sin} ^ 3 \theta$

$= \frac{1}{256} \cdot \frac{1}{\frac{x}{\sqrt{16 + {x}^{2}}}} - \frac{1}{768} \cdot \frac{1}{{x}^{3} / {\left({x}^{2} + 16\right)}^{\frac{3}{2}}} + C$

$= \frac{\sqrt{16 + {x}^{2}}}{256 x} - {\left({x}^{2} + 16\right)}^{\frac{3}{2}} / \left(768 {x}^{3}\right) + C$