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

$\textcolor{b l u e}{\int {e}^{x} / \sqrt{{e}^{2 x} + 4} \mathrm{dx} = \ln \left(\sqrt{{e}^{2 x} + 4} + {e}^{x}\right) + C}$

#### Explanation:

Integration by Trigonometric Substitution

We let ${e}^{x} = 2 \cdot \tan \theta$
Let ${e}^{2 x} = 4 \cdot {\tan}^{2} \theta$

Let ${e}^{x} \mathrm{dx} = 2 \cdot {\sec}^{2} \theta \cdot d \theta$
and $\mathrm{dx} = \frac{2 \cdot {\sec}^{2} \theta \cdot d \theta}{e} ^ x = \frac{2 \cdot {\sec}^{2} \theta \cdot d \theta}{2 \cdot \tan \theta} = \frac{{\sec}^{2} \theta \cdot d \theta}{\tan \theta}$

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

$\int {e}^{x} / \sqrt{{e}^{2 x} + 4} \mathrm{dx} = \int \frac{2 \cdot \tan \theta}{\sqrt{4} \sqrt{{\tan}^{2} \theta + 1}} \frac{{\sec}^{2} \theta \cdot d \theta}{\tan \theta}$

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

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

$\int {e}^{x} / \sqrt{{e}^{2 x} + 4} \mathrm{dx} = \int \frac{\cancel{2} \cdot \cancel{\tan} \theta}{\cancel{2} \cdot \cancel{\sec} \theta} \frac{\cancel{\sec} \theta \cdot \sec \theta \cdot d \theta}{\cancel{\tan} \theta}$

$\int {e}^{x} / \sqrt{{e}^{2 x} + 4} \mathrm{dx} = \int \sec \theta \cdot d \theta = \ln \left(\sec \theta + \tan \theta\right) + {C}_{1}$

From ${e}^{x} = 2 \cdot \tan \theta$
${e}^{x}$ is Opposite side
$2$ is Adjacent side
$\sqrt{{e}^{2 x} + 4}$ is Hypotenuse

$\int {e}^{x} / \sqrt{{e}^{2 x} + 4} \mathrm{dx} = \int \sec \theta \cdot d \theta = \ln \left(\sec \theta + \tan \theta\right) + C$

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

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

$\int {e}^{x} / \sqrt{{e}^{2 x} + 4} \mathrm{dx} = \ln \left(\sqrt{{e}^{2 x} + 4} + {e}^{x}\right) - \ln 2 + {C}_{1}$

$\textcolor{b l u e}{\int {e}^{x} / \sqrt{{e}^{2 x} + 4} \mathrm{dx} = \ln \left(\sqrt{{e}^{2 x} + 4} + {e}^{x}\right) + C}$

God bless....I hope the explanation is useful.