## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{2} \frac{d x}{\sqrt{\left(16-x^{2}\right)^{3}}}
$$
## Solution

$$
\int_{0}^{2} \frac{d x}{\sqrt{\left(16-x^{2}\right)^{3}}}=
$$

Substitution:

$$
\begin{aligned}
& x=4 \sin t ; d x=4 \cos t d t \\
& x=0 \Rightarrow t=\arcsin \frac{0}{4}=0 \\
& x=2 \Rightarrow t=\arcsin \frac{2}{4}=\arcsin \frac{1}{2}=\frac{\pi}{6}
\end{aligned}
$$

We get:

$$
\begin{aligned}
& =\int_{0}^{\pi / 6} \frac{4 \cos t d t}{\sqrt{\left(16-16 \sin ^{2} t\right)^{3}}}=\int_{0}^{\pi / 6} \frac{4 \cos t d t}{4^{3} \sqrt{\left(1-\sin ^{2} t\right)^{3}}}=\frac{1}{16} \int_{0}^{\pi / 6} \frac{\cos t d t}{\cos ^{3} t}=\frac{1}{16} \int_{0}^{\pi / 6} \frac{d t}{\cos ^{2} t}= \\
& =\left.\frac{1}{16} \tan t\right|_{0} ^{\pi / 6}=\frac{1}{16} \tan \frac{\pi}{6}-\frac{1}{16} \tan 0=\frac{1}{16} \cdot \frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{48}
\end{aligned}
$$

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 $\% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+12-22$ » Categories: Kuznetsov's Problem Book Integrals Problem 12 | Integrals

- Last edited: 15:55, 5 July 2009.
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## Problem Kuznetsov Integrals 12-23

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