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

Apr 9, 2018

$\int \frac{x}{\sqrt{16 + {x}^{2}}} \mathrm{dx} = \ln | \sqrt{{x}^{2} + 16} + x | + C$

#### Explanation:

For integrals involving the root $\sqrt{{x}^{2} + {a}^{2}} ,$ we can use the substitution $x = a \tan \theta$. Here, ${a}^{2} = 16 , a = 4 , x = 4 \tan \theta$

So, $\mathrm{dx} = 4 {\sec}^{2} \theta d \theta$ and we can apply the substitution.

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

$= \int {\sec}^{2} \frac{\theta}{\sqrt{1 + {\tan}^{2} \theta}} d \theta$

Recalling the identity $1 + {\tan}^{2} \theta = {\sec}^{2} \theta ,$

$= \int {\sec}^{2} \frac{\theta}{\sqrt{{\sec}^{2} \theta}} d \theta$

$= \int {\sec}^{\cancel{2}} \frac{\theta}{\cancel{\sec \theta d}} \theta$

$= \int \sec \theta d \theta = \ln | \sec \theta + \tan \theta | + C$

This is a fairly common integral, it should be memorized. We need things in terms of $x .$ Recalling that $x = 4 \tan \theta , \tan \theta = \frac{x}{4.}$

We still need the secant in terms of $x .$ Applying the identity $1 + {\tan}^{2} \theta = {\sec}^{2} \theta :$

$\frac{16}{16} + {x}^{2} / 16 = {\sec}^{2} \theta$

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

Thus, we have

$\ln | \frac{\sqrt{{x}^{2} + 16} + x}{4} | + C = \ln | \sqrt{{x}^{2} + 16} + x | - \ln \left(4\right) + C$

We may absorb the $\ln 4$ into the constant of integration. We're left with

$\int \frac{x}{\sqrt{16 + {x}^{2}}} \mathrm{dx} = \ln | \sqrt{{x}^{2} + 16} + x | + C$