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

Sep 10, 2016

This cannot be integrated using elementary functions.

#### Explanation:

Use the substitution ${x}^{2} = 4 \tan \theta$. This implies that $2 x \mathrm{dx} = 4 {\sec}^{2} \theta d \theta$. Also keep in mind that $x = 2 \sqrt{\tan} \theta$.

We have:

$\int {x}^{2} / \sqrt{16 + {x}^{4}} \mathrm{dx} = \frac{1}{2} \int \frac{x \left(2 x \mathrm{dx}\right)}{\sqrt{16 + {x}^{4}}}$

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

Note that $1 + {\tan}^{2} \theta = {\sec}^{2} \theta$, so $\sec \theta = \sqrt{1 + {\tan}^{2} \theta}$:

$= \int \frac{\sqrt{\tan} \theta \left({\sec}^{2} \theta\right) d \theta}{\sec} \theta = \int \sqrt{\tan} \theta \sec \theta d \theta$

The more we continue, we see that this cannot be integrated using elementary functions.