## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{4 \sqrt{3}} \frac{d x}{\sqrt{\left(64-x^{2}\right)^{3}}}
$$
## Solution

$$
\int_{0}^{4 \sqrt{3}} \frac{d x}{\sqrt{\left(64-x^{2}\right)^{3}}}=
$$

Substitution:

$$
\begin{aligned}
& x=8 \sin t ; d x=8 \cos t d t \\
& x=0 \Rightarrow t=\arcsin \frac{0}{8}=0 \\
& x=4 \sqrt{3} \Rightarrow t=\arcsin \frac{4 \sqrt{3}}{8}=\frac{\pi}{3}
\end{aligned}
$$

We get:

$$
\begin{aligned}
& =\int_{0}^{\pi / 3} \frac{8 \cos t d t}{\sqrt{\left(64-64 \sin ^{2} t\right)^{3}}}=\int_{0}^{\pi / 3} \frac{8 \cos t d t}{2^{9} \sqrt{\left(1-\sin ^{2} t\right)^{3}}}=\frac{1}{2^{6}} \int_{0}^{\pi / 3} \frac{\cos t d t}{\cos ^{3} t}=\frac{1}{64} \int_{0}^{\pi / 3} \frac{d t}{\cos ^{2} t}= \\
& =\left.\frac{1}{64} \operatorname{tg} t\right|_{0} ^{\pi / 3}=\frac{1}{64} \operatorname{tg} \frac{\pi}{3}-\frac{1}{64} \operatorname{tg} 0=\frac{\sqrt{3}}{64}
\end{aligned}
$$

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 $\% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+12-17$ » Categories: Kuznetsov's Problem Book Integrals Problem $12 \mid$ Integrals

- Last edited: 13:41, 5 July 2009.
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## Problem Kuznetsov Integrals 12-18

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