# How do you integrate int 1/sqrt(e^(2x)+12e^x+45)dx using trigonometric substitution?

Jul 14, 2018

$\frac{1}{3 \sqrt{5}} \ln | \tan \left[\frac{1}{2} \left\{\arctan \left(\frac{{e}^{x} - 6}{3}\right) - \arctan 2\right\}\right] | + C$.

#### Explanation:

Let, I=int1/sqrt(e^(2x)+12e^x+45)dx .

${e}^{2 x} + 12 {e}^{x} + 45 = {\left({e}^{x} + 6\right)}^{2} + {3}^{2}$.

So, let us subst. ${e}^{x} + 6 = 3 \tan y , \text{ so that, } {e}^{x} \mathrm{dx} = 3 {\sec}^{2} y \mathrm{dy}$.

$\therefore \mathrm{dx} = \frac{3 {\sec}^{2} y \mathrm{dy}}{e} ^ x = \frac{3 {\sec}^{2} y \mathrm{dy}}{3 \tan y - 6} = \frac{3 {\sec}^{2} y \mathrm{dy}}{3 \left(\tan y - 2\right)}$.

$\therefore I = \int \frac{1}{\sqrt{{\left(3 \tan y\right)}^{2} + {3}^{2}}} \cdot \frac{3 {\sec}^{2} y \mathrm{dy}}{3 \left(\tan y - 2\right)}$,

$= \frac{1}{3} \int \frac{\sec y \mathrm{dy}}{\tan y - 2}$,

$\therefore I = \frac{1}{3} \int \frac{\mathrm{dy}}{\sin y - 2 \cos y}$.

Now, $\sin y - 2 \cos y = \sqrt{5} \cdot \left(\frac{1}{\sqrt{5}} \sin y - \frac{2}{\sqrt{5}} \cos y\right)$.

$\text{So, if we set "cosalpha=1/sqrt5," then, } \sin \alpha = \frac{2}{\sqrt{5}} , \mathmr{and} ,$

$\sin y - 2 \cos y = \sqrt{5} \left(\sin y \cos \alpha - \cos y \sin \alpha\right) = \sqrt{5} \sin \left(y - \alpha\right)$.

$\therefore I = \frac{1}{3} \int \frac{1}{\sqrt{5} \sin \left(y - \alpha\right)} \mathrm{dy}$,

$= \frac{1}{3 \sqrt{5}} \int \csc \left(y - \alpha\right) \mathrm{dy}$,

$= \frac{1}{3 \sqrt{5}} \ln | \tan \left(\frac{y - \alpha}{2}\right) |$,

$\Rightarrow I = \frac{1}{3 \sqrt{5}} \ln | \tan \left[\frac{1}{2} \left\{\arctan \left(\frac{{e}^{x} - 6}{3}\right) - \arctan 2\right\}\right] | + C$.

Jul 14, 2018

$- \frac{1}{3 \sqrt{5}} \ln | \left\{\frac{15 t + 2}{\sqrt{5}} + \sqrt{45 {t}^{2} + 12 t + 1}\right\} | + C$,

where, $t = \frac{1}{y - 6}$.

#### Explanation:

Here is a way to solve the Problem without using trigo.

substn.

Let, $I = \int \frac{1}{\sqrt{{e}^{2 x} + 12 {e}^{x} + 45}} \mathrm{dx} = \int \frac{1}{\sqrt{{\left({e}^{x} + 6\right)}^{2} + {3}^{2}}} \mathrm{dx} .$

Letting $\left({e}^{x} + 6\right) = y , {e}^{x} \mathrm{dx} = \mathrm{dy} , \mathmr{and} , \mathrm{dx} = \frac{\mathrm{dy}}{e} ^ x = \frac{\mathrm{dy}}{y - 6}$.

$\therefore I = \int \frac{1}{\left(y - 6\right) \sqrt{{y}^{2} + 9}} \mathrm{dy}$.

Next, we take,

$\left(y - 6\right) = \frac{1}{t} . \therefore \mathrm{dy} = - \frac{1}{t} ^ 2 \mathrm{dt} . \text{ Also, } y = 6 + \frac{1}{t} .$

$\therefore I = \int \frac{1}{\frac{1}{t} \sqrt{{\left(6 + \frac{1}{t}\right)}^{2} + 9}} \left(- \frac{1}{t} ^ 2\right) \mathrm{dt}$,

$= - \int \frac{1}{\sqrt{{\left(6 t + 1\right)}^{2} + 9 {t}^{2}}} \mathrm{dt}$,

$= - \int \frac{1}{\sqrt{45 {t}^{2} + 12 t + 1}} \mathrm{dt}$,

$= - \int \frac{1}{\sqrt{\left(3 \sqrt{5} t + \frac{2}{\sqrt{5}}\right) + {\left(\frac{1}{\sqrt{5}}\right)}^{2}}} \mathrm{dt}$,

$= - \frac{1}{3 \sqrt{5}} \ln | \left\{\left(3 \sqrt{5} t + \frac{2}{\sqrt{5}}\right) + \sqrt{45 {t}^{2} + 12 t + 1}\right\} |$,

$= - \frac{1}{3 \sqrt{5}} \ln | \left\{\frac{15 t + 2}{\sqrt{5}} + \sqrt{45 {t}^{2} + 12 t + 1}\right\} | + C$,

where, $t = \frac{1}{y - 6}$.