## Task Condition

Calculate the definite integral:

$$
\int_{\arccos (1 / \sqrt{10})}^{\arccos (1 / \sqrt{26})} \frac{12 d x}{(6+5 \tan x) \sin 2 x}
$$
## Solution

Let's use the substitution:

$$
t=\operatorname{tg} x
$$

From which:

$$
\begin{aligned}
& \sin 2 x=\frac{2 t}{1+t^{2}}, d x=\frac{d t}{1+t^{2}} \\
& x=\arccos \frac{1}{\sqrt{10}} \Rightarrow t=\operatorname{tg}\left(\arccos \frac{1}{\sqrt{10}}\right)=\sqrt{\frac{1}{\cos ^{2}\left(\arccos \frac{1}{\sqrt{10}}\right)}-1}= \\
& =\sqrt{\frac{1}{\left(\frac{1}{10}\right)}-1}=\sqrt{10-1}=\sqrt{9}=3 \\
& x=\arccos \frac{1}{\sqrt{26}} \Rightarrow t=\operatorname{tg}\left(\arccos \frac{1}{\sqrt{26}}\right)=\sqrt{\frac{1}{\cos ^{2}\left(\arccos \frac{1}{\sqrt{26}}\right)}-1}= \\
& =\sqrt{\frac{1}{\left(\frac{1}{26}\right)}-1}=\sqrt{26-1}=\sqrt{25}=5
\end{aligned}
$$

Substitute:

$$
\begin{aligned}
& \quad \int_{\arccos (1 / \sqrt{10})}^{\arccos (1 / \sqrt{26})} \frac{12 d x}{(6+5 \operatorname{tg} x) \sin 2 x}=\int_{3}^{5} \frac{12}{(6+5 t) \cdot \frac{2 t}{1+t^{2}}} \cdot \frac{d t}{1+t^{2}}= \\
& =\int_{3}^{5} \frac{6}{t(6+5 t)} d t=
\end{aligned}
$$

Decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:

$$
\begin{aligned}
& \frac{6 d t}{t(6+5 t)}=\frac{A}{t}+\frac{B}{6+5 t}=\frac{A(6+5 t)+B t}{t(6+5 t)}=\frac{(5 A+B) t+6 A}{t(6+5 t)} \\
& \left\{\begin{array}{l}
5 A+B=0 \\
6 A=6
\end{array}\right. \\
& \left\{\begin{array}{l}
B=-5 \\
A=1
\end{array}\right. \\
& \frac{6 d t}{t(6+5 t)}=\frac{1}{t}-\frac{5}{6+5 t}
\end{aligned}
$$

We get:

$$
\begin{aligned}
& =\int_{3}^{5}\left(\frac{1}{t}-\frac{5}{6+5 t}\right) d t=\int_{3}^{5} \frac{d t}{t}-\int_{3}^{5} \frac{d(5 t+6)}{6+5 t}= \\
& =\ln \left|t \left\|_{3}^{5}-\ln \left|5 t+6 \|_{3}^{5}=\ln \right| 5|-\ln | 3|-\ln | 5 \cdot 5+6|+\ln | 5 \cdot 3+6 \mid=\right.\right. \\
& =\ln \frac{5}{3}-\ln 31+\ln 21=\ln \frac{5 \cdot 21}{3 \cdot 31}=\ln \frac{105}{93}
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+9-29$ » Categories: Kuznetsov's Problem Book Integrals Problem 9| Integrals

Ukrainian Banner Network

- Last edited: 17:11, 14 May 2009.
- Content is available under CC-BY-SA 3.0.

Created by Gee Teatoo

## Problem Kuznetsov Integrals 9-30

## Material from PlusPi