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

Mar 15, 2018

$I = \frac{1}{2} x \sqrt{{x}^{2} + 16} - \ln | \frac{x + \sqrt{{x}^{2} + 16}}{2} | + C$

#### Explanation:

Since the square root in the denominator is of the form $\sqrt{{a}^{2} + {x}^{2}}$, we will use the substitution $x = a \tan \theta = 4 \tan \theta$.

Therefore, $\mathrm{dx} = 4 {\sec}^{2} \theta d \theta$.

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

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

$I = \int \frac{16 {\tan}^{2} \theta}{\sqrt{16 {\sec}^{2} \theta}} 4 {\sec}^{2} \theta$

$I = \int \frac{16 {\tan}^{2} \theta}{4 \sec \theta} \cdot 4 {\sec}^{2} \theta$

$I = \int 16 {\tan}^{2} \theta \sec \theta$

$I = \int 16 \left({\sec}^{2} \theta - 1\right) \sec \theta$

$I = 16 \int {\sec}^{3} \theta - \sec \theta$

These are two known integrals.

$I = 8 \sec \theta \tan \theta - 8 \ln | \tan \theta + \sec \theta | + C$

Now recall that $\frac{x}{4} = \tan \theta$, therefore $\sec \theta = \frac{\sqrt{{x}^{2} + 16}}{4}$.

$I = \frac{1}{2} x \sqrt{{x}^{2} + 16} - 8 \ln | \frac{x + \sqrt{{x}^{2} + 16}}{2} | + C$

Hopefully this helps!