# How do you integrate int 1/sqrt(-e^(2x) +81)dx using trigonometric substitution?

$\int \frac{1}{\sqrt{- {e}^{2 x} + 81}}$ = $\frac{1}{9} \ln \left(9 - \sqrt{81 - {e}^{2 x}}\right) - \frac{x}{9} + C$

#### Explanation:

Use the Trigonometric substitution

${e}^{x} = 9 \cdot \sin \theta$

${e}^{x}$ $\mathrm{dx}$ = $9 \cdot \cos \theta$ $d$$\theta$

${e}^{2 x} = 81 {\sin}^{2} \theta$

$\mathrm{dx}$=(9cos theta)/(9sin theta $d$$\theta$

$\mathrm{dx}$=$\cos \frac{\theta}{\sin} \theta$ $d$$\theta$

$\int \frac{1}{\sqrt{81 - {e}^{2 x}}}$$\mathrm{dx}$

$\int \frac{1}{\sqrt{81 - 81 {\sin}^{2} \theta}}$$\cos \frac{\theta}{\sin} \theta$ $d$$\theta$

$\int \frac{1}{\sqrt{81} \cdot \sqrt{1 - {\sin}^{2} \theta}}$$\cos \frac{\theta}{\sin} \theta$ $d$$\theta$

$\int \frac{1}{9 \sqrt{{\cos}^{2} \theta}}$$\cos \frac{\theta}{\sin} \theta$ $d$$\theta$

$\frac{1}{9} \cdot \int \csc \theta$ $d$$\theta$

$\frac{1}{9} \cdot \ln \left(\csc \theta - \cot \theta\right) + C$

$\frac{1}{9} \cdot \ln \left(\frac{9}{e} ^ x - \frac{\sqrt{81 - {e}^{2 x}}}{e} ^ x\right) + C$

$\frac{1}{9} \cdot \ln \left(\frac{9 - \sqrt{81 - {e}^{2 x}}}{e} ^ x\right) + C$

$\frac{1}{9} \cdot \left(\ln \left(9 - \sqrt{81 - {e}^{2 x}}\right) - \ln {e}^{x}\right) + C$

$\frac{1}{9} \cdot \left(\ln \left(9 - \sqrt{81 - {e}^{2 x}}\right) - x\right) + C$

$\frac{1}{9} \cdot \ln \left(9 - \sqrt{81 - {e}^{2 x}}\right) - \frac{x}{9} + C$