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

Jul 25, 2016

$\int \frac{d x}{\sqrt{{x}^{2} + 4}} = l n \left(\frac{x}{2} + \sqrt{1 + {x}^{2} / 4}\right) + C$

#### Explanation:

int (d x)/(sqrt (x^2+4))=?

$\text{let be "u=x/2" ; " d x=2*d u" ; } {x}^{2} = 4 {u}^{2}$

int (2*d u)/(sqrt(4u^2+4))=int(2*d u)/(sqrt(4(u^2+1))

$\int \frac{\cancel{2} \cdot d u}{\cancel{2} \cdot \sqrt{{u}^{2} + 1}} = \int \frac{d u}{\sqrt{{u}^{2} + 1}}$

$\text{now, substitute "u=tan v" ; } v = a r c \tan u$

$d u = {\sec}^{2} v \cdot d v$

$\int \frac{{\sec}^{2} v \cdot d v}{\sqrt{{\tan}^{2} v + 1}} \text{ ;so } {\tan}^{2} v + 1 = {\sec}^{2} v$

$\int \frac{{\sec}^{2} v \cdot d v}{\sqrt{{\sec}^{2} v}} = \int \frac{{\cancel{\sec}}^{2} v \cdot d v}{\cancel{\sec} v} = \int \sec v \cdot d v$

$\text{expand fraction by } \tan v + \sec v$

$\int \sec v \cdot d v \cdot \frac{\tan v + \sec v}{\tan v + \sec v}$

$\int \frac{\sec v \cdot \tan v + {\sec}^{2} v}{\tan v + \sec v} \cdot d v$

$k = \tan v + \sec v$

$d k = \left(\sec v \cdot \tan v + {\sec}^{2} v\right) \cdot d v$

$\int \frac{d k}{k} = l n k + C$

$\text{undo substitution } k = \tan v + \sec v$

$\int \frac{d x}{\sqrt{{x}^{2} + 4}} = l n \left(\tan v + \sec v\right) + C$

$\sec v = \sqrt{1 + {\tan}^{2} v} = \sqrt{1 + {u}^{2}}$

$\int \frac{d x}{\sqrt{{x}^{2} + 4}} = l n \left(u + \sqrt{1 + {u}^{2}}\right) + C$

$\text{undo substitution } u = \frac{x}{2}$

$\int \frac{d x}{\sqrt{{x}^{2} + 4}} = l n \left(\frac{x}{2} + \sqrt{1 + {x}^{2} / 4}\right) + C$