## Problem Statement

Calculate the definite integral:

$$
\int_{3}^{6} \frac{\sqrt{x^{2}-9}}{x^{4}} d x
$$
## Solution

$$
\int_{3}^{6} \frac{\sqrt{x^{2}-9}}{x^{4}} d x=
$$

Substitution:

$$
\begin{aligned}
& x=\frac{3}{\sin t} \Rightarrow d x=-\frac{3}{\sin ^{2} t} \cdot \cos t d t=-\frac{3 \cos t d t}{\sin ^{2} t} \\
& x=3 \Rightarrow t=\arcsin \frac{3}{3}=\frac{\pi}{2} \\
& x=6 \Rightarrow t=\arcsin \frac{3}{6}=\frac{\pi}{6}
\end{aligned}
$$

We get:

$$
\begin{aligned}
& =\int_{\pi / 2}^{\pi / 6} \frac{\sqrt{\frac{9}{\sin ^{2} t}-9}}{\left(\frac{3}{\sin t}\right)^{4}} \cdot\left(-\frac{3 \cos t d t}{\sin ^{2} t}\right)=\int_{\pi / 6}^{\pi / 2} \frac{\sin ^{4} t}{3^{4}} \cdot 3 \sqrt{\frac{1-\sin ^{2} t}{\sin ^{2} t}} \cdot \frac{3 \cos t d t}{\sin ^{2} t}= \\
& =\frac{1}{9} \cdot \int_{\pi / 6}^{\pi / 2} \sin ^{2} t \cdot \sqrt{\frac{\cos ^{2} t}{\sin ^{2} t}} \cdot \cos t d t=\frac{1}{9} \cdot \int_{\pi / 6}^{\pi / 2} \sin ^{2} t \cdot \frac{\cos t}{\sin t} \cdot \cos t d t= \\
& =\frac{1}{9} \cdot \int_{\pi / 6}^{\pi / 2} \sin t \cdot \cos ^{2} t d t=-\frac{1}{9} \cdot \int_{\pi / 6}^{\pi / 2} \cos ^{2} t d(\cos t)=-\left.\frac{\cos ^{3} t}{27}\right|_{\pi / 6} ^{\pi / 2}= \\
& =-\frac{\cos ^{3} \frac{\pi}{2}}{27}+\frac{\cos ^{3} \frac{\pi}{6}}{27}=-\frac{0^{3}}{27}+\frac{\left(\frac{\sqrt{3}}{2}\right)^{3}}{27}=0+\frac{\left(\frac{3 \sqrt{3}}{8}\right)}{27}=\frac{\sqrt{3}}{72}
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/%D0%9A%D1%83%D0%B7%D0%BD%D0%B5%D1%86%D0%BE%D0%B2_%D0%98%D0%BD%D1%82%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D1%8B_12-24»

Categories: Kuznetsov's Problem Book Integrals Problem 12 | Integrals

- Last modified: 16:05, 5 July 2009.
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Created by GeeTeatoo

## Problem Kuznetsov Integrals 12-25

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