## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{5} \frac{d x}{\left(25+x^{2}\right) \sqrt{25+x^{2}}}
$$
## Solution

$$
\int_{0}^{5} \frac{d x}{\left(25+x^{2}\right) \sqrt{25+x^{2}}}=\int_{0}^{5} \frac{d x}{\left(25+x^{2}\right)^{3 / 2}}=
$$

Substitution:

$$
\begin{aligned}
& x=5 \operatorname{tg} t \Rightarrow d x=\frac{5 d t}{\cos ^{2} t} \\
& x=0 \Rightarrow t=\operatorname{arctg} \frac{0}{5}=0 \\
& x=5 \Rightarrow t=\operatorname{arctg} \frac{5}{5}=\frac{\pi}{4}
\end{aligned}
$$

We get:

$$
\begin{aligned}
& =\int_{0}^{\pi / 4} \frac{5 d t}{\left(25+25 \operatorname{tg}^{2} t\right)^{3 / 2} \cdot \cos ^{2} t}=\int_{0}^{\pi / 4} \frac{5 d t}{5^{3}\left(1+\operatorname{tg}^{2} t\right)^{3 / 2} \cdot \cos ^{2} t}= \\
& =\int_{0}^{\pi / 4} \frac{d t}{25\left(\frac{1}{\cos ^{2} t}\right)^{3 / 2} \cdot \cos ^{2} t}=\frac{1}{25} \int_{0}^{\pi / 4} \frac{d t}{\frac{1}{\cos ^{3} t} \cdot \cos ^{2} t}=\frac{1}{25} \int_{0}^{\pi / 4} \cos t d t= \\
& =\left.\frac{1}{25} \sin t\right|_{0} ^{\pi / 4}=\frac{1}{25} \sin \frac{\pi}{4}-\frac{1}{25} \sin 0=\frac{1}{25} \cdot \frac{\sqrt{2}}{2}-0=\frac{\sqrt{2}}{50}
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_12-3$ » Categories: Kuznetsov's Problem Book Integrals Problem 12 | Integrals

- Last edited: 13:53, 5 July 2009.
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Kuznetsov's Problem Book Integrals 12-3 — PlusPi

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