# How do you integrate int 1/sqrt(-e^(2x)-12e^x-35)dx using trigonometric substitution?

##### 1 Answer
May 13, 2018

$I = \frac{2}{\sqrt{35}} {\tan}^{-} 1 \left(\frac{\sqrt{7 \left(- {e}^{x} - 5\right)}}{\sqrt{5 \left({e}^{x} + 7\right)}}\right) + c$

#### Explanation:

Here,

$I = \int \frac{1}{\sqrt{- {e}^{2 x} - 12 {e}^{x} - 35}} \mathrm{dx}$

Now,

$- {e}^{2 x} - 12 {e}^{x} - 35 = 1 - \left({e}^{2 x} + 12 {e}^{x} + 36\right) = 1 - {\left({e}^{x} + 6\right)}^{2}$

$\therefore I = \int \frac{1}{\sqrt{1 - {\left({e}^{x} + 6\right)}^{2}}} \mathrm{dx}$

Subst. ${e}^{x} + 6 = \cos u \implies {e}^{x} = \cos u - 6 \implies {e}^{x} \mathrm{dx} = - \sin u \mathrm{du}$

$\implies \mathrm{dx} = - \sin \frac{u}{e} ^ x \mathrm{du} = - \sin \frac{u}{\cos u - 6} \mathrm{du} = \sin \frac{u}{6 - \cos u} \mathrm{du}$

$I = \int \frac{1}{\sqrt{1 - {\cos}^{2} u}} \times \sin \frac{u}{6 - \cos u} \mathrm{du}$

$= \int \frac{1}{\sin} u \times \sin \frac{u}{6 - \cos u} \mathrm{du}$

$= \int \frac{1}{6 - \cos u} \mathrm{du}$

Again subst. $\tan \left(\frac{u}{2}\right) = t \implies {\sec}^{2} \left(\frac{u}{2}\right) \cdot \frac{1}{2} \mathrm{du} = \mathrm{dt}$

=>du=(2dt)/(1+tan^2(u/2))=(2dt)/(1+t^2)and cosu=(1- t^2)/(1+t^2)
So,$I = \int \frac{1}{6 - \frac{1 - {t}^{2}}{1 + {t}^{2}}} \times \left(\frac{2}{1 + {t}^{2}}\right) \mathrm{dt}$

$= \int \frac{2}{6 + 6 {t}^{2} - 1 + {t}^{2}} \mathrm{dt}$

$= \int \frac{2}{7 {t}^{2} + 5} \mathrm{dt}$

$= \frac{2}{7} \int \frac{1}{{t}^{2} + {\left(\sqrt{\frac{5}{7}}\right)}^{2}} \mathrm{dt}$

$= \frac{2}{7} \times \frac{1}{\sqrt{\frac{5}{7}}} {\tan}^{-} 1 \left(\frac{t}{\sqrt{\frac{5}{7}}}\right) + c$

$= \frac{2}{\sqrt{35}} {\tan}^{-} 1 \left(\frac{\sqrt{7} t}{\sqrt{5}}\right) + c , w h e r e , t = \tan \left(\frac{u}{2}\right)$

$\therefore I = \frac{2}{\sqrt{35}} {\tan}^{-} 1 \left(\frac{\sqrt{7} \tan \left(\frac{u}{2}\right)}{\sqrt{5}}\right) + c$

Now,

${\tan}^{2} \left(\frac{u}{2}\right) = \frac{1 - \cos u}{1 + \cos u} , w h e r e , \cos u = {e}^{x} + 6$

:.tan^2(u/2)=(1-(e^x+6))/(1+(e^x+6))=(-e^x-5)/(e^x+7) <0 ..??

=>tan(u/2)=sqrt((-e^x-5)/(e^x+7)
Thus,

$I = \frac{2}{\sqrt{35}} {\tan}^{-} 1 \left(\frac{\sqrt{7 \left(- {e}^{x} - 5\right)}}{\sqrt{5 \left({e}^{x} + 7\right)}}\right) + c$