## Problem Statement

Calculate the definite integral:

$$
\int_{1}^{2} \frac{\sqrt{x^{2}-1}}{x^{4}} d x
$$
## Solution

$$
\int_{1}^{2} \frac{\sqrt{x^{2}-1}}{x^{4}} d x=
$$

Substitution:

$$
\begin{aligned}
& x=\frac{1}{\sin t} \Rightarrow d x=-\frac{1}{\sin ^{2} t} \cdot \cos t d t=-\frac{\cos t d t}{\sin ^{2} t} \\
& x=1 \Rightarrow t=\arcsin \frac{1}{1}=\frac{\pi}{2} \\
& x=2 \Rightarrow t=\arcsin \frac{1}{2}=\frac{\pi}{6}
\end{aligned}
$$

We get:

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

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_12-6»

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

- Last modified: 14:21, 5 July 2009.
- Content is available under CC-BY-SA 3.0.

Created by GeeTeatoo

## Problem Kuznetsov Integrals 12-7

## Material from PlusPi