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

Jul 25, 2018

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

#### Explanation:

Trigonometric substitution usually is needed when the function contains a term of ${x}^{2} \pm {a}^{2}$ or ${a}^{2} \pm {x}^{2}$.

Let's convert it to such a form by substituting $u = {e}^{x}$ and $\mathrm{du} = {e}^{x} \mathrm{dx}$:

$\int \setminus \frac{1}{\sqrt{{u}^{2} + 36}} \setminus \mathrm{du}$

Now, we need to decide which trigonometric function to substitute in this integral. The options are $u = \left\{\begin{matrix}a \sin \left(\theta\right) \\ a \tan \left(\theta\right) \\ a \sec \left(\theta\right)\end{matrix}\right.$, where $a$ is some constant. It should be apparent by trial substitution that only $u = 6 \tan \left(\theta\right)$ will simplify the integral.

Thus, substitute $u = 6 \tan \left(\theta\right)$ and $\mathrm{du} = 6 {\sec}^{2} \left(\theta\right) d \theta$ to get

$6 \int \setminus {\sec}^{2} \frac{\theta}{\sqrt{36 {\tan}^{2} \left(\theta\right) + 36}} \setminus d \theta$

Pull out the $36$ from the denominator (which cancels the $6$ on the outside):

$\int \setminus {\sec}^{2} \frac{\theta}{\sqrt{{\tan}^{2} \left(\theta\right) + 1}} \setminus d \theta$

Now, the reason we could substitute $x = a \tan \left(\theta\right)$ in the first place was so that we could take advantage of the identity ${\tan}^{2} \left(\theta\right) + 1 = {\sec}^{2} \left(\theta\right)$.

This simplifies nicely to

$\int \setminus \sec \left(\theta\right) \setminus d \theta$

Now, to finish this problem, you can recall the integral of the secant function as $\ln | \sec \left(\theta\right) + \tan \left(\theta\right) | + C$ (which is probably given to you for tests that have formula booklets). Then, substitute back $u = 6 \tan \left(\theta\right)$ and $u = {e}^{x}$ to get the final answer of

$\ln | \frac{1}{6} \left(\sqrt{{u}^{2} + 36} + u\right) | + C$
$= \ln | \frac{1}{6} \left(\sqrt{{e}^{2 x} + 36} + {e}^{x}\right) | + C$
$= \ln | \sqrt{{e}^{2 x} + 36} + {e}^{x} | - 6 + C$
$= \ln | \sqrt{{e}^{2 x} + 36} + {e}^{x} | + C$

(Note how in the final line we absorbed the $6$ into the arbitrary constant $C$.)

$- - - - - - - - - - - - - - - - - - - -$

Here I will give you a quick derivation for the integral of $\sec \left(\theta\right)$.

The trick here is to multiply it by $\frac{\sec \left(\theta\right) + \tan \left(\theta\right)}{\sec \left(\theta\right) + \tan \left(\theta\right)}$:

$\int \setminus \sec \left(\theta\right) \cdot \frac{\sec \left(\theta\right) + \tan \left(\theta\right)}{\sec \left(\theta\right) + \tan \left(\theta\right)} \setminus d \theta$

$= \int \setminus \frac{\sec \left(\theta\right) \tan \left(\theta\right) + {\sec}^{2} \left(\theta\right)}{\sec \left(\theta\right) + \tan \left(\theta\right)} \setminus d \theta$

This may seem at first like needlessly complicating the problem, but in fact now you can simplify this problem by substituting $u = \sec \left(\theta\right) + \tan \left(\theta\right)$ and $\mathrm{du} = \left(\sec \left(\theta\right) \tan \left(\theta\right) + {\sec}^{2} \left(\theta\right)\right) d \theta$:

$\int \setminus \frac{\mathrm{du}}{u}$

$= \ln | u | + C$

$= \ln | \sec \left(\theta\right) + \tan \left(\theta\right) | + C$