## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{\sqrt{5} / 2} \frac{d x}{\sqrt{\left(5-x^{2}\right)^{3}}}
$$
## Solution

$$
\int_{0}^{\sqrt{5} / 2} \frac{d x}{\sqrt{\left(5-x^{2}\right)^{3}}}=
$$

Substitution:

$$
\begin{aligned}
& x=\sqrt{5} \sin t ; d x=\sqrt{5} \cos t d t \\
& x=0 \Rightarrow t=\arcsin \frac{0}{\sqrt{5}}=0 \\
& x=\frac{\sqrt{5}}{2} \Rightarrow t=\arcsin \frac{\left(\frac{\sqrt{5}}{2}\right)}{\sqrt{5}}=\arcsin \frac{1}{2}=\frac{\pi}{6}
\end{aligned}
$$

We get:

$$
\begin{aligned}
& =\int_{0}^{\pi / 6} \frac{\sqrt{5} \cos t d t}{\sqrt{\left(5-5 \sin ^{2} t\right)^{3}}}=\int_{0}^{\pi / 6} \frac{\sqrt{5} \cos t d t}{\sqrt{5^{3}} \sqrt{\left(1-\sin ^{2} t\right)^{3}}}=\frac{1}{5} \int_{0}^{\pi / 6} \frac{\cos t d t}{\cos ^{3} t}=\frac{1}{5} \int_{0}^{\pi / 6} \frac{d t}{\cos ^{2} t}= \\
& =\left.\frac{1}{5} \operatorname{tg} t\right|_{0} ^{\pi / 6}=\frac{1}{5} \operatorname{tg} \frac{\pi}{6}-\frac{1}{5} \operatorname{tg} 0=\frac{1}{5} \cdot \frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{15}
\end{aligned}
$$

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}+12-5 »

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

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## Problem Kuznetsov Integrals 12-6

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