# How do you integrate int (e^x-1)/sqrt(e^(2x) -16)dx using trigonometric substitution?

Jul 26, 2018

$- i \left(\arcsin \left({e}^{x} / 4\right) + \frac{1}{4} \ln \left(| \frac{4 + \sqrt{16 - {e}^{2 x}}}{e} ^ x |\right)\right) + c$

#### Explanation:

e^x=4sinΘ
x=ln(4sinΘ)
dx=1/(4sinΘ)*4cosΘdΘ
dx=cosΘ/sinΘdΘ

int(e^x-1)/(sqrt(e^(2x)-16)$\mathrm{dx}$

int(4sinΘ-1)/(sqrt((4sinΘ)^2-16)*cosΘ/sinΘdΘ

int(4sinΘ-1)/(sqrt(16sin^2Θ-16))*cosΘ/sinΘdΘ

int(4sinΘ-1)/(sqrt(16)*sqrt(sin^2Θ-1)*cosΘ/sinΘdΘ

int(4sinΘ-1)/(4*sqrt(-cos^2Θ)*cosΘ/sinΘdΘ

1/iint(4sinΘ-1)/(4cosΘ*cosΘ/sinΘdΘ

-iint(4sinΘ-1)/(4sinΘ)dΘ

-iint((4sinΘ)/(4sinΘ)-(1)/(4sinΘ))dΘ

-iint1dΘ+iint1/4cscΘdΘ

-iΘ + i/4ln|cscΘ+cotΘ| + c

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{- i \arcsin \left({e}^{x} / 4\right) - \frac{i}{4} \ln \left(| \frac{4 + \sqrt{16 - {e}^{2 x}}}{e} ^ x |\right) + c , c \in \mathbb{R}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Note 1: We set ${e}^{x}$ as a form of sinΘ because the ${e}^{x}$ term in the root in the denominator was positive and the constant term was negative.

Note 2: intcscΘ=$- \ln | \csc x + \cot x | + c$

Note 3: You get the expression for Θ from rewriting the original equation, and you can solve for cscΘ and cotΘ using a right triangle.

Note 4: There are probably ways to simplify my final answer, but I just kept it like that.

Aug 6, 2018

$I = \int \frac{{e}^{x} - 1}{\sqrt{{e}^{2 x} - 16}} \mathrm{dx}$
Let $X = {e}^{x}$

$x = \ln \left(X\right)$

$\mathrm{dx} = \frac{\mathrm{dX}}{X}$

So:

$I = \int \frac{X - 1}{X \sqrt{{X}^{2} - 16}} \mathrm{dX}$

$= \int \frac{\cancel{X}}{\cancel{X} \sqrt{{X}^{2} - 16}} \mathrm{dX} - \int \frac{1}{X \sqrt{{X}^{2} - 16}} \mathrm{dX}$

Now let $X = 4 \sec \left(\theta\right)$

$\mathrm{dX} = 4 \sec \left(\theta\right) \tan \left(\theta\right) d \theta$

So :

$I = \int \frac{4 \sec \left(\theta\right) \tan \left(\theta\right)}{\sqrt{{\left(4 \sec \left(\theta\right)\right)}^{2} - 16}} d \theta - \int \frac{\cancel{4 \sec \left(\theta\right)} \tan \left(\theta\right)}{\cancel{4 \sec \left(\theta\right)} \sqrt{{\left(4 \sec \left(\theta\right)\right)}^{2} - 16}} d \theta$

$= \int \frac{4 \sec \left(\theta\right) \tan \left(\theta\right)}{\sqrt{16 \left(\sec {\left(\theta\right)}^{2} - 1\right)}} d \theta - \int \frac{\tan \left(\theta\right)}{\sqrt{16 \left(\sec {\left(\theta\right)}^{2} - 1\right)}} d \theta$

Because $\sec {\left(\theta\right)}^{2} - 1 = \tan {\left(\theta\right)}^{2}$,

$I = \int \frac{\cancel{4} \sec \left(\theta\right) \cancel{\tan \left(\theta\right)}}{\cancel{4 \tan \left(\theta\right)}} d \theta - \int \frac{\cancel{\tan \left(\theta\right)}}{4 \cancel{\tan \left(\theta\right)}} d \theta$

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

Now, we have $\int \sec \left(\theta\right) d \theta = \ln \left(| \sec \left(\theta\right) + \tan \left(\theta\right) |\right)$

Here is a proof if you need the explanation.

So :
$I = \ln \left(| \sec \left(\theta\right) + \tan \left(\theta\right) |\right) - \frac{1}{4} \theta$

Because $\theta = {\sec}^{- 1} \left(\frac{X}{4}\right) = {\sec}^{- 1} \left({e}^{x} / 4\right)$, and $\tan \left({\sec}^{- 1} \left(u\right)\right) = \sqrt{{u}^{2} - 1}$, where $u$ is a function (here is a proof if you need it.),

$I = \ln \left(| {e}^{x} / 4 + \sqrt{{e}^{2 x} / 16 - 1} |\right) - \frac{1}{4} {\sec}^{- 1} \left({e}^{x} / 4\right) + C$, $C \in \mathbb{R}$.