# How do you integrate int sqrt(e^(8x)-9) using trig substitutions?

Sep 15, 2016

$\frac{\sqrt{{e}^{8 x} - 9} - 3 \text{arcsec} \left({e}^{4 x} / 3\right)}{4} + C$

#### Explanation:

$\int \sqrt{{e}^{8 x} - 9} \mathrm{dx}$

Apply the substitution ${e}^{4 x} = 3 \sec \theta$. Thus $4 {e}^{4 x} \mathrm{dx} = 3 \sec \theta \tan \theta d \theta$.

Note that $\mathrm{dx} = \frac{3 \sec \theta \tan \theta d \theta}{4 {e}^{4 x}} = \frac{3 \sec \theta \tan \theta d \theta}{4 \left(3 \sec \theta\right)} = \frac{1}{4} \tan \theta d \theta$.

$= \int \sqrt{{\left({e}^{4 x}\right)}^{2} - 9} \mathrm{dx}$

$= \int \sqrt{9 {\sec}^{2} \theta - 9} \left(\frac{1}{4} \tan \theta d \theta\right)$

$= \frac{3}{4} \int \sqrt{{\sec}^{2} \theta - 1} \left(\tan \theta d \theta\right)$

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

$= \frac{3}{4} \int {\tan}^{2} \theta d \theta$

$= \frac{3}{4} \int \left({\sec}^{2} \theta - 1\right) d \theta$

$= \frac{3}{4} \int {\sec}^{2} \theta d \theta - \frac{3}{4} \int d \theta$

$= \frac{3}{4} \tan \theta - \frac{3}{4} \theta + C$

From ${e}^{4 x} = 3 \sec \theta$, we see that $\theta = \text{arcsec} \left({e}^{4 x} / 3\right)$. Also, since $\sec \theta = {e}^{4 x} / 3$, we see that the hypotenuse is ${e}^{4 x}$, the adjacent side is $3$, and the opposite side is $\sqrt{{e}^{8 x} - 9}$. Thus $\tan \theta = \frac{\sqrt{{e}^{8 x} - 9}}{3}$.

$= \frac{3}{4} \left(\frac{\sqrt{{e}^{8 x} - 9}}{3}\right) - \frac{3}{4} \text{arcsec} \left({e}^{4 x} / 3\right) + C$

$= \frac{\sqrt{{e}^{8 x} - 9} - 3 \text{arcsec} \left({e}^{4 x} / 3\right)}{4} + C$