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

Mar 29, 2018

need double check. I'm not absolute sure..but I hope this helps

Explanation:

First of all we need to rearrange the expresion under square root.

Lets do $z = {e}^{x}$. With this we have

sqrt(-z^2+12z-45)=sqrt(-(z^2-12z+45). Now lets complete the square under square root

$\sqrt{- \left({\left(z - 6\right)}^{2} - 36 + 45\right)} = \sqrt{- \left({\left(z - 6\right)}^{2} + 9\right)}$

Now lets do the variable change $z - 6 = 3 \tan r$. With this change we have

$\sqrt{- \left({\left(z - 6\right)}^{2} + 9\right)} = \sqrt{- \left(9 {\tan}^{2} r + 9\right)} = \sqrt{- 9} \sqrt{{\tan}^{2} r + 1}$

and $\mathrm{dz} = {\sec}^{2} r \mathrm{dr}$

We know that $\sqrt{- 9} = 3 i$ and ${\tan}^{2} r + 1 = {\sec}^{2} r$

Now we have finally

$\int \frac{1}{3 i \sec r} {\sec}^{2} r \mathrm{dr} = \int \frac{1}{3 i} \sec r \mathrm{dr} = \frac{1}{3 i} \int \sec r \mathrm{dr}$

Mar 29, 2018

$\int \setminus \frac{1}{\sqrt{- {e}^{2 x} + 12 {e}^{x} - 45}} \setminus \mathrm{dx} = \frac{i}{3 \sqrt{5}} \ln | \frac{6 \left(\sqrt{{\left({e}^{x} - 6\right)}^{2} / 9 + 1} - 1\right)}{{e}^{x} - 6} - \sqrt{5} - 1 | - \frac{i}{3 \sqrt{5}} \ln | \frac{6 \left(\sqrt{{\left({e}^{x} - 6\right)}^{2} / 9 + 1} - 1\right)}{{e}^{x} - 6} + \sqrt{5} - 1 | + C$

Explanation:

We seek:

$I = \int \setminus \frac{1}{\sqrt{- {e}^{2 x} + 12 {e}^{x} - 45}} \setminus \mathrm{dx}$

The denominator is quadratic in ${e}^{x}$, so can complete the square to get:

$I = \int \setminus \frac{1}{\sqrt{\left(- 1\right) \left({e}^{2 x} - 12 {e}^{x} + 45\right)}} \setminus \mathrm{dx}$
$\setminus \setminus = \int \setminus \frac{1}{\sqrt{\left(- 1\right) \left({\left({e}^{x} - 6\right)}^{2} - {6}^{2} + 45\right)}} \setminus \mathrm{dx}$
$\setminus \setminus = \int \setminus \frac{1}{\sqrt{- 1} \sqrt{{\left({e}^{x} - 6\right)}^{2} - 36 + 45}} \setminus \mathrm{dx}$
$\setminus \setminus = \int \setminus \frac{1}{\sqrt{{\left({e}^{x} - 6\right)}^{2} + 9}} \cdot \frac{1}{i} \mathrm{dx}$
$\setminus \setminus = \int \setminus \frac{1}{\sqrt{{\left({e}^{x} - 6\right)}^{2} + 9}} \cdot \frac{1}{i} \cdot \frac{i}{i} \mathrm{dx}$
$\setminus \setminus = - i \setminus \int \setminus \frac{1}{\sqrt{{\left({e}^{x} - 6\right)}^{2} + 9}} \mathrm{dx}$
$\setminus \setminus = - i \setminus {I}_{1}$, where ${I}_{1} = \int \setminus \frac{1}{\sqrt{{\left({e}^{x} - 6\right)}^{2} + 9}} \mathrm{dx}$

Note that the denominator is positive $\forall x \in \mathbb{R}$, so we do not get a real solution.

To proceed with the integral, ${I}_{1}$, we can perform a substitution, Let:

${e}^{x} - 6 = 3 \tan \theta \implies {e}^{x} = 3 \tan \theta + 6$

And if we differentiate wrt $x$ we get:

${e}^{x} = 3 {\sec}^{2} \theta \frac{d \theta}{\mathrm{dx}} \implies \frac{d \theta}{\mathrm{dx}} = \frac{3 \tan \theta + 6}{3 {\sec}^{2} \theta}$

And substituting into the integral ${I}_{1}$ we get:

${I}_{1} = \int \setminus \frac{1}{\sqrt{{\left(3 \tan \theta\right)}^{2} + 9}} \frac{3 {\sec}^{2} \theta}{3 \tan \theta + 6} \setminus d \theta$

$\setminus \setminus \setminus = \int \setminus \frac{1}{\sqrt{9 \left({\tan}^{2} \theta + 1\right)}} \frac{3 {\sec}^{2} \theta}{3 \left(\tan \theta + 2\right)} \setminus d \theta$

$\setminus \setminus \setminus = \int \setminus \frac{1}{\sqrt{9 \left({\sec}^{2} \theta\right)}} \frac{{\sec}^{2} \theta}{\left(\tan \theta + 2\right)} \setminus d \theta$

$\setminus \setminus \setminus = \frac{1}{3} \setminus \int \setminus \frac{\sec \theta}{\tan \theta + 2} \setminus d \theta$

Next, we can perform a tangent half angle (Weierstrass substitution), so that:

$t = \tan \left(\frac{\theta}{2}\right) \implies \theta = 2 \arctan t$

And we find that:

$\sin \theta = \frac{2 t}{1 + {t}^{2}}$, $\cos \theta = \frac{1 - {t}^{2}}{1 + {t}^{2}}$ and $\frac{d \theta}{\mathrm{dt}} = \frac{2}{1 + {t}^{2}}$

Thus we can perform a second substitution:

${I}_{1} = \frac{1}{3} \setminus \int \setminus \frac{\frac{1}{\cos} \theta}{\left(\sin \frac{\theta}{\cos} \theta\right) + 2} \setminus d \theta$

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

$\setminus \setminus \setminus = \frac{1}{3} \setminus \int \setminus \frac{\frac{2}{1 - {t}^{2}}}{\frac{2 t}{1 - {t}^{2}} + 2} \setminus \mathrm{dt}$

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

$\setminus \setminus \setminus = - \frac{1}{3} \setminus \int \setminus \frac{1}{{t}^{2} - t - 1} \setminus \mathrm{dt}$

This integrand, is now significantly simplified, and we can use partial fraction decomposition (skipped) toi write in the form:

${I}_{1} = - \frac{1}{3} \frac{2}{\sqrt{5}} \left\{\int \frac{1}{2 t - \sqrt{5} - 1} - \frac{1}{2 t + \sqrt{5} - 1} \setminus \mathrm{dt}\right\}$

Which we can now evaluate, to get:

${I}_{1} = - \frac{1}{3 \sqrt{5}} \left\{\ln | 2 t - \sqrt{5} - 1 | - \ln | 2 t + \sqrt{5} - 1 |\right\}$

We can then restore the $t$ substitution, using $t = \tan \left(\frac{\theta}{2}\right)$, to get:

${I}_{1} = - \frac{1}{3 \sqrt{5}} \left\{\ln | 2 \tan \left(\frac{\theta}{2}\right) - \sqrt{5} - 1 | - \ln | 2 \tan \left(\frac{\theta}{2}\right) + \sqrt{5} - 1 |\right\}$

And we can restore the $\theta$ substitution using the relationship, , $\theta = \arctan \left(\frac{1}{3} \left({e}^{x} - 6\right)\right)$ from which we get:

$\tan \left(\frac{\theta}{2}\right) = \frac{3 \left(\sqrt{{\left({e}^{x} - 6\right)}^{2} / 9 + 1} - 1\right)}{{e}^{x} - 6}$

And so we get:

${I}_{1} = - \frac{1}{3 \sqrt{5}} \left\{\ln | \frac{6 \left(\sqrt{{\left({e}^{x} - 6\right)}^{2} / 9 + 1} - 1\right)}{{e}^{x} - 6} - \sqrt{5} - 1 | - \ln | \frac{6 \left(\sqrt{{\left({e}^{x} - 6\right)}^{2} / 9 + 1} - 1\right)}{{e}^{x} - 6} + \sqrt{5} - 1 |\right\}$

Thus:

$I = \frac{i}{3 \sqrt{5}} \ln | \frac{6 \left(\sqrt{{\left({e}^{x} - 6\right)}^{2} / 9 + 1} - 1\right)}{{e}^{x} - 6} - \sqrt{5} - 1 | - \frac{i}{3 \sqrt{5}} \ln | \frac{6 \left(\sqrt{{\left({e}^{x} - 6\right)}^{2} / 9 + 1} - 1\right)}{{e}^{x} - 6} + \sqrt{5} - 1 | + C$