# How do you integrate int 1/sqrt(-e^(2x) +9)dx using trigonometric substitution?

Mar 14, 2016

$\int \left(\frac{1}{\sqrt{- {e}^{2 x} + 9}}\right) = \frac{\ln \left(\sqrt{9 - {e}^{2 x}} - 3\right) - x}{3}$

#### Explanation:

Here is the corresponding right triangle from which you can extract all the parts for substitution: You need to find the following and then systematically replace all x terms (rectangular) with $\theta$ terms (trig):

x

$\frac{\mathrm{dx}}{d \theta}$

$\sqrt{9 - {e}^{2 x}}$

I'll start you off in the right direction:

Since

sin(theta) =e^x/3

We have then that:

$x = \ln \left(3 \cdot \sin \left(\theta\right)\right)$

Now differentiate with respect to $\theta$ and split up the differentials.

You can then rewrite that entire integrand in terms of $\theta$ and solve using trigonometric substitution.

I'll give you one more piece and then I gotta sleep!

What is cos(theta)?

Well, referencing the triangle you can see that:

$\cos \left(\theta\right) = \frac{\sqrt{9 - {e}^{2 x}}}{3}$

This says that:

$\sqrt{9 - {e}^{2 x}} = 3 \cdot \cos \left(\theta\right)$

You'll still need to find $d \theta$

So now you can systematically replace all rectangular terms with
trigonometric terms.