## Problem Statement

Calculate the definite integral:

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

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

Substitution:

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

We get:

$$
\begin{aligned}
& =\int_{\pi / 2}^{\pi / 6} \frac{\sqrt{\frac{2}{\sin ^{2} t}-2}}{\left(\frac{\sqrt{2}}{\sin t}\right)^{4}} \cdot\left(-\frac{\sqrt{2} \cos t d t}{\sin ^{2} t}\right)=\int_{\pi / 6}^{\pi / 2} \frac{\sin ^{4} t}{4} \cdot \sqrt{2} \sqrt{\frac{1-\sin ^{2} t}{\sin ^{2} t}} \cdot \frac{\sqrt{2} \cos t d t}{\sin ^{2} t}= \\
& =\frac{1}{2} \int_{\pi / 6}^{\pi / 2} \sin ^{2} t \cdot \sqrt{\frac{\cos ^{2} t}{\sin ^{2} t}} \cdot \cos t d t=\frac{1}{2} \int_{\pi / 6}^{\pi / 2} \sin ^{2} t \cdot \frac{\cos t}{\sin t} \cdot \cos t d t= \\
& =\frac{1}{2} \int_{\pi / 6}^{\pi / 2} \sin t \cdot \cos ^{2} t d t=-\frac{1}{2} \int_{\pi / 6}^{\pi / 2} \cos ^{2} t d(\cos t)=-\left.\frac{\cos ^{3} t}{6}\right|_{\pi / 6} ^{\pi / 2}= \\
& =-\frac{\cos ^{3} \frac{\pi}{2}}{6}+\frac{\cos ^{3} \frac{\pi}{6}}{6}=-\frac{0^{3}}{6}+\frac{\left(\frac{\sqrt{3}}{2}\right)^{3}}{6}=0+\frac{\left(\frac{3 \sqrt{3}}{8}\right)}{6}=\frac{\sqrt{3}}{16}
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+12-18$ »

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

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## Problem Kuznetsov Integrals 12-19

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