# How do you find the integral of int 1/(t^2-9)^(1/2)dt from 4 to 6?

May 9, 2018

${\int}_{4}^{6} \setminus \frac{1}{\sqrt{{t}^{2} - 9}} \setminus \mathrm{dt} = \ln \left(\frac{6 + \sqrt{27}}{4 + \sqrt{7}}\right) \approx 0.5215924 \ldots$

#### Explanation:

We seek:

$I = {\int}_{4}^{6} \setminus \frac{1}{\sqrt{{t}^{2} - 9}} \setminus \mathrm{dt}$

We can perform a trigonometric substitution, Let

$t = 3 \sec \theta \implies \frac{\mathrm{dt}}{d \theta} = 3 \sec \theta \tan \theta$

We would normally change the limits of integration from $x$ to $\theta$, however let us omit this step and consider the corresponding indefinite integral, which after substitution we get:

$\int \setminus \frac{1}{\sqrt{{t}^{2} - 9}} \setminus \mathrm{dt} = \int \setminus \frac{1}{\sqrt{9 {\sec}^{2} \theta - 9}} 3 \sec \theta \tan \theta \setminus d \theta$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \int \setminus \frac{1}{3 \sqrt{{\sec}^{2} \theta - 1}} 3 \sec \theta \tan \theta \setminus d \theta$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \int \setminus \frac{1}{\sqrt{{\tan}^{2} \theta}} \sec \theta \tan \theta \setminus d \theta$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \int \setminus \sec \theta \setminus d \theta$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \ln | \sec \theta + \tan \theta | + C$

And if we restore the substitution we get:

$\int \setminus \frac{1}{\sqrt{{t}^{2} - 9}} \setminus \mathrm{dt} = \ln | \frac{t}{3} + \sqrt{{\left(\frac{t}{3}\right)}^{2} - 1} | + C$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \ln | \frac{1}{3} \left(t + \sqrt{{t}^{2} - 9}\right) | + C$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \ln | t + \sqrt{{t}^{2} - 9} | + \ln \left(\frac{1}{3}\right) + C$

$I = {\left[\setminus \setminus \ln | t + \sqrt{{t}^{2} - 9} | \setminus \setminus\right]}_{4}^{6}$

$\setminus \setminus = \ln | 6 + \sqrt{36 - 9} | - \ln | 4 + \sqrt{16 - 9} |$

$\setminus \setminus = \ln \left(6 + \sqrt{27}\right) - \ln \left(4 + \sqrt{7}\right)$

$\setminus \setminus = \ln \left(\frac{6 + \sqrt{27}}{4 + \sqrt{7}}\right)$

$\setminus \setminus \approx 0.5215924 \ldots$

May 9, 2018

$I = \ln \left(\frac{6 + \sqrt{27}}{4 + \sqrt{7}}\right)$

#### Explanation:

We know that,

color(red)(int1/sqrt(x^2-a^2)dx=ln|x+sqrt(x^2-a^2)|+c

Here,

$I = {\int}_{4}^{6} \frac{1}{\sqrt{{t}^{2} - 9}} \mathrm{dt}$

$= {\int}_{4}^{6} \frac{1}{\sqrt{{t}^{2} - {\left(3\right)}^{2}}} \mathrm{dt}$

=color(red)([ln|t+sqrt(t^2-3^2)|]_4^6

$= \ln | 6 + \sqrt{{6}^{2} - {3}^{2}} | - \ln | 4 + \sqrt{{4}^{2} - {3}^{2}} |$

$= \ln | 6 + \sqrt{27} | - \ln | 4 + \sqrt{7} |$

$I = \ln \left(\frac{6 + \sqrt{27}}{4 + \sqrt{7}}\right)$

From the graph ,we can say that $f \left(t\right) = \frac{1}{\sqrt{{t}^{2} - 9}}$
is continuous on[4,6]

graph{1/sqrt(x^2-9) [-3.77, 6.23, -0.7, 4.3]}