# Evaluate the integral? :  int \ x^2/(sqrt(x^2-25))^5 \ dx

Jul 17, 2017

$\int {x}^{2} / {\left(\sqrt{{x}^{2} - 25}\right)}^{5} \mathrm{dx} = - \frac{1}{75} {x}^{3} / {\left(\sqrt{{x}^{2} - 25}\right)}^{3} + C$

#### Explanation:

$\int {x}^{2} / {\left(\sqrt{{x}^{2} - 25}\right)}^{5} \mathrm{dx} = \int {x}^{2} / {\left(5 \sqrt{{\left(\frac{x}{5}\right)}^{2} - 1}\right)}^{5} \mathrm{dx}$

$\int {x}^{2} / {\left(\sqrt{{x}^{2} - 25}\right)}^{5} \mathrm{dx} = \frac{1}{5} ^ 5 \int {x}^{2} / {\left(\sqrt{{\left(\frac{x}{5}\right)}^{2} - 1}\right)}^{5} \mathrm{dx}$

$\int {x}^{2} / {\left(\sqrt{{x}^{2} - 25}\right)}^{5} \mathrm{dx} = \frac{1}{25} \int {\left(\frac{x}{5}\right)}^{2} / {\left(\sqrt{{\left(\frac{x}{5}\right)}^{2} - 1}\right)}^{5} d \left(\frac{x}{5}\right)$

Substitute:

$\frac{x}{5} = \sec t$

$d \left(\frac{x}{5}\right) = \sec t \tan t$

to have:

$\int {x}^{2} / {\left(\sqrt{{x}^{2} - 25}\right)}^{5} \mathrm{dx} = \frac{1}{25} \int {\sec}^{2} \frac{t}{\sqrt{{\sec}^{2} t - 1}} ^ 5 \sec t \tan t \mathrm{dt}$

use now the trigonometric identity:

${\sec}^{2} t - 1 = {\tan}^{2} t$

$\int {x}^{2} / {\left(\sqrt{{x}^{2} - 25}\right)}^{5} \mathrm{dx} = \frac{1}{25} \int {\sec}^{2} \frac{t}{\sqrt{{\tan}^{2} t}} ^ 5 \sec t \tan t \mathrm{dt}$

If we restrict ourselves to the interval $x \in \left(5 , + \infty\right)$ so that $t \in \left(0 , \frac{\pi}{2}\right)$ we have that $\tan t > 0$ so $\sqrt{{\tan}^{2} t} = \tan t$:

$\int {x}^{2} / {\left(\sqrt{{x}^{2} - 25}\right)}^{5} \mathrm{dx} = \frac{1}{25} \int {\sec}^{2} \frac{t}{\tan} ^ 5 t \sec t \tan t \mathrm{dt}$

$\int {x}^{2} / {\left(\sqrt{{x}^{2} - 25}\right)}^{5} \mathrm{dx} = \frac{1}{25} \int {\sec}^{3} \frac{t}{\tan} ^ 4 t \mathrm{dt}$

$\int {x}^{2} / {\left(\sqrt{{x}^{2} - 25}\right)}^{5} \mathrm{dx} = \frac{1}{25} \int \frac{1}{\cos} ^ 3 t {\cos}^{4} \frac{t}{\sin} ^ 4 t \mathrm{dt}$

$\int {x}^{2} / {\left(\sqrt{{x}^{2} - 25}\right)}^{5} \mathrm{dx} = \frac{1}{25} \int \cos \frac{t}{\sin} ^ 4 t \mathrm{dt}$

$\int {x}^{2} / {\left(\sqrt{{x}^{2} - 25}\right)}^{5} \mathrm{dx} = \frac{1}{25} \int \frac{d \left(\sin t\right)}{\sin} ^ 4 t$

$\int {x}^{2} / {\left(\sqrt{{x}^{2} - 25}\right)}^{5} \mathrm{dx} = - \frac{1}{75} \frac{1}{\sin} ^ 3 t + C$

To undo the substitution note that:

$\sin t = \sqrt{1 - {\cos}^{2} t} = \sqrt{1 - \frac{1}{\sec} ^ 2 t} = \sqrt{1 - \frac{25}{x} ^ 2} = \frac{\sqrt{{x}^{2} - 25}}{x}$

Then:

$\int {x}^{2} / {\left(\sqrt{{x}^{2} - 25}\right)}^{5} \mathrm{dx} = - \frac{1}{75} {x}^{3} / {\left(\sqrt{{x}^{2} - 25}\right)}^{3} + C$

Now consider $x \in \left(- \infty , - 5\right)$ and substitute $t = - x$:

$\int {x}^{2} / {\left(\sqrt{{x}^{2} - 25}\right)}^{5} \mathrm{dx} = \int {\left(- t\right)}^{2} / {\left(\sqrt{{\left(- t\right)}^{2} - 25}\right)}^{5} d \left(- t\right)$

$\int {x}^{2} / {\left(\sqrt{{x}^{2} - 25}\right)}^{5} \mathrm{dx} = - \int {t}^{2} / {\left(\sqrt{{t}^{2} - 25}\right)}^{5} \mathrm{dt}$

$\int {x}^{2} / {\left(\sqrt{{x}^{2} - 25}\right)}^{5} \mathrm{dx} = \frac{1}{75} {t}^{3} / {\left(\sqrt{{t}^{2} - 25}\right)}^{3} + C$

and undoing the substitution:

$\int {x}^{2} / {\left(\sqrt{{x}^{2} - 25}\right)}^{5} \mathrm{dx} = - \frac{1}{75} {x}^{3} / {\left(\sqrt{{x}^{2} - 25}\right)}^{3} + C$

so the expression is the same for $x \in \left(- \infty , 5\right) \cup \left(5 , + \infty\right)$

Jul 17, 2017

$\int \setminus {x}^{2} / {\left(\sqrt{{x}^{2} - 25}\right)}^{5} \setminus \mathrm{dx} = - \frac{1}{75} \setminus {x}^{3} / {\left(\sqrt{{x}^{2} - 25}\right)}^{3} + C$

#### Explanation:

Compare the denominator to the trig identity:

${\tan}^{2} A \equiv {\sec}^{2} A - 1$

In an attempt to reduce the denominator to something simper.

Why this identity in particular? Well firstly it is a trig identity of the form $f n {1}^{2} = f n {2}^{2} - 1$ so with an appropriate factor it will reduce to being dependant upon $f n 1$ alone, and secondly, $\frac{d}{\mathrm{dx}} \tan x = {\sec}^{2} x$ and $\frac{d}{\mathrm{dx}} \sec = \sec x \tan x$ so either derivative also appears in the identity, which should help with a substitution in the integral,

So we want to find:

$I = \int \setminus {x}^{2} / {\left(\sqrt{{x}^{2} - 25}\right)}^{5} \setminus \mathrm{dx}$
$\setminus \setminus = \int \setminus {x}^{2} / {\left(\sqrt{25 \left({x}^{2} / 25 - 1\right)}\right)}^{5} \setminus \mathrm{dx}$
$\setminus \setminus = \int \setminus {x}^{2} / {\left(5 \sqrt{{\left(\frac{x}{5}\right)}^{2} - 1}\right)}^{5} \setminus \mathrm{dx}$

So let us try a substitution $\sec \theta = \frac{x}{5}$ such that

$x = 5 \sec \theta$, and $\frac{\mathrm{dx}}{d \theta} = 5 \sec \theta \tan \theta$

Thus:

$I = \int \setminus {\left(5 \sec \theta\right)}^{2} / {\left(5 \left(\sqrt{{\sec}^{2} \theta - 1}\right)\right)}^{5} 5 \sec \theta \tan \theta \setminus d \theta$

$\setminus \setminus = \int \setminus \frac{{5}^{3} {\sec}^{3} \theta \tan \theta}{5 \left(\sqrt{{\tan}^{2} \theta}\right)} ^ 5 \setminus d \theta$

$\setminus \setminus = \int \setminus \frac{{5}^{3} {\sec}^{3} \theta \tan \theta}{{5}^{5} {\tan}^{5} \theta} \setminus d \theta$
$\setminus \setminus = \frac{1}{25} \setminus \int \setminus \frac{{\sec}^{3} \theta}{{\tan}^{4} \theta} \setminus d \theta$

$\setminus \setminus = \frac{1}{25} \setminus \int \setminus \frac{\frac{1}{\cos} ^ 3 \theta}{{\sin}^{4} \frac{\theta}{\cos} ^ 4 \theta} \setminus d \theta$

$\setminus \setminus = \frac{1}{25} \setminus \int \setminus \frac{\cos \theta}{{\sin}^{4} \theta} \setminus d \theta$

Which we can integrate by observation as:

$\frac{d}{\mathrm{du}} \frac{1}{\sin} ^ 3 u = - 3 \cos \frac{u}{\sin} ^ 4 u$

And so we conclude that:

$I = \frac{\frac{1}{25} \setminus \frac{1}{\sin} ^ 3 \theta}{- 3} + C$
$\setminus \setminus = - \frac{1}{75} \setminus \frac{1}{\sin} ^ 3 \theta + C$

If we refer back to our earlier substitution, we note:

$x = 5 \sec \theta \implies \sec \theta = \frac{x}{5}$
$\cos \theta = \frac{5}{x}$

And using the trig identity ${\cos}^{2} A + {\sin}^{2} A \equiv 1$ this gives:

${\sin}^{2} \theta + {\left(\frac{5}{x}\right)}^{2} = 1 \implies \sin \theta = \sqrt{1 - \frac{25}{x} ^ 2}$

Using this to reverse the earlier substitution we get:

$I = - \frac{1}{75} \setminus \frac{1}{\sqrt{1 - \frac{25}{x} ^ 2}} ^ 3 + C$
$\setminus \setminus = - \frac{1}{75} \setminus \frac{1}{\sqrt{1 - \frac{25}{x} ^ 2}} ^ 3 + C$
$\setminus \setminus = - \frac{1}{75} \setminus \frac{1}{\sqrt{\frac{1}{x} ^ 2 \left({x}^{2} - 25\right)}} ^ 3 + C$
$\setminus \setminus = - \frac{1}{75} \setminus \frac{1}{\frac{1}{x} \sqrt{{x}^{2} - 25}} ^ 3 + C$
$\setminus \setminus = - \frac{1}{75} \setminus {x}^{3} / {\left(\sqrt{{x}^{2} - 25}\right)}^{3} + C$
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 58, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}