# How do you integrate int e^(2x)/sqrt(e^(2x) -36)dx using trigonometric substitution?

Mar 10, 2016

$\sqrt{{e}^{2 x} - 36} + C$

#### Explanation:

We don't really need substitution since we have a perfect form
$\int \frac{f ' \left(x\right)}{2 \sqrt{f \left(x\right)}} \to \sqrt{f \left(x\right)}$
but here you go:

• ${e}^{x} = 6 \cosh u$
• ${e}^{x} \mathrm{dx} = \left(6 \sinh u\right) \mathrm{du}$

Hence:

${e}^{2 x} - 36 = {\left(6 \cosh u\right)}^{2} - 36 = 36 \left({\cosh}^{2} u - 1\right)$
$36 {\sinh}^{2} u = {\left(6 \sinh u\right)}^{2}$

So we have:

$\int {e}^{2 x} / \sqrt{{e}^{2 x} - 36} \mathrm{dx} = \int {e}^{x} / \sqrt{{e}^{2 x} - 36} \left({e}^{x} \mathrm{dx}\right) = \int \frac{6 \cosh u}{6 \sinh u} \left(6 \sinh u \mathrm{du}\right) = \int 6 \cosh u \mathrm{du} = 6 \sinh u + C$

To substitute back we use:
$\sinh u = \sqrt{{\cosh}^{2} u - 1} = \sqrt{{e}^{2 x} / 36 - 1} = \frac{\sqrt{{e}^{2 x} - 36}}{6}$

Finally:

$\int {e}^{2 x} / \sqrt{{e}^{2 x} - 36} \mathrm{dx} = \sqrt{{e}^{2 x} - 36} + C$