# How do you integrate int 1/sqrt(e^(2x)+12e^x-13)dx using trigonometric substitution?

May 11, 2018

$I = \frac{2}{\sqrt{13}} {\tan}^{-} 1 \left(\sqrt{13} \times \sqrt{\frac{{e}^{x} - 1}{{e}^{x} + 13}}\right) + c$

#### Explanation:

Here,

$I = \int \frac{1}{\sqrt{{e}^{2 x} + 12 {e}^{x} - 13}} \mathrm{dx}$

$= \int \frac{1}{\sqrt{{e}^{2 x} + 12 {e}^{x} + 36 - 49}} \mathrm{dx}$

$= \int \frac{1}{\sqrt{{\left({e}^{x} + 6\right)}^{2} - {7}^{2}}} \mathrm{dx}$

Let,

color(blue)(e^x+6=7secu=>e^x=7secu-6=>e^xdx=7secutanudu

color(blue)(=>dx=(7secutanudu)/(7secu-6)&,secu= (e^x+6)/7=>color(red)(cosu=7/(e^x+6)

$\therefore I = \int \frac{1}{\sqrt{{7}^{2} {\sec}^{2} u - {7}^{2}}} \times \frac{7 \sec u \tan u}{7 \sec u - 6} \mathrm{du}$

$= \int \frac{1}{7 \tan u} \times \frac{7 \sec u \tan u}{7 \sec u - 6} \mathrm{du}$

$= \int \sec \frac{u}{7 \sec u - 6} \mathrm{du}$

$= \int \frac{1}{7 - 6 \cos u} \mathrm{du}$

Now,subst. color(violet)(tan(u/2)=t=>sec^2(u/2)*1/2du=dt

color(violet)(=>du=(2dt)/(1+tan^2(u/2))=(2dt)/(1+t^2) and cosx=(1- t^2)/(1+t^2)

:.I=int1/(7-6((1-t^2)/(1+t^2)))xx2/(1+t^2)dt=2intdt/(7+7t^2- 6+6t^2)

$= 2 \int \frac{1}{13 {t}^{2} + 1} \mathrm{dt} = \frac{2}{13} \int \frac{1}{{t}^{2} + \left(\frac{1}{13}\right)} \mathrm{dt}$

$= \frac{2}{13} \int \frac{1}{{t}^{2} + {\left(\frac{1}{\sqrt{13}}\right)}^{2}} \mathrm{dt}$

$= \frac{2}{13} \times \frac{1}{\frac{1}{\sqrt{13}}} {\tan}^{-} 1 \left(\frac{t}{\frac{1}{\sqrt{13}}}\right) + c$

=2/sqrt13tan^-1(sqrt13t)+c, where,color(violet)(t=tan(u/2)

$\therefore I = \frac{2}{\sqrt{13}} {\tan}^{-} 1 \left(\sqrt{13} \tan \left(\frac{u}{2}\right)\right) + c \ldots \to \left(A\right)$

We know that, tan^2(u/2)=(1-cosu)/(1+cosu),where, color(red)(cosu=7/(e^x+6)

=>tan^2(u/2)=(1-(7/(e^x+6)))/(1+(7/(e^x+6)))=(e^x+6- 7)/(e^x+6+7)=(e^x-1)/(e^x+13)

$\implies \tan \left(\frac{u}{2}\right) = \sqrt{\frac{{e}^{x} - 1}{{e}^{x} + 13}}$

Hence, from $\left(A\right)$, we have

$I = \frac{2}{\sqrt{13}} {\tan}^{-} 1 \left(\sqrt{13} \times \sqrt{\frac{{e}^{x} - 1}{{e}^{x} + 13}}\right) + c$