## Solution

Let's use the substitution:

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

From which we get:

$$
\begin{aligned}
& \sin 2 x=\frac{2 t}{1+t^{2}}, d x=\frac{d t}{1+t^{2}} \\
& x=\frac{\pi}{4} \Rightarrow t=\operatorname{tg} \frac{\pi}{4}=1 \\
& x=\arccos \frac{1}{\sqrt{26}} \Rightarrow t=\operatorname{tg}\left(\arccos \frac{1}{\sqrt{26}}\right)=\sqrt{\frac{1}{\cos ^{2}\left(\arccos \frac{1}{\sqrt{26}}\right)}-1}= \\
& =\sqrt{\frac{1}{\left(\frac{1}{26}\right)}-1}=\sqrt{26-1}=\sqrt{25}=5
\end{aligned}
$$

Substitute:

$$
\begin{aligned}
& \quad \int_{\pi / 4}^{\arccos (1 / \sqrt{26})} \frac{36 d x}{(6-\operatorname{tg} x) \sin 2 x}=36 \cdot \int_{1}^{5} \frac{1}{(6-t) \cdot \frac{2 t}{1+t^{2}}} \cdot \frac{d t}{1+t^{2}}= \\
& =36 \cdot \int_{1}^{5} \frac{1}{2 t(6-t)} d t=-\frac{36}{2} \cdot \int_{1}^{5} \frac{1}{t(t-6)} d t=
\end{aligned}
$$

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

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

We get:

$$
\begin{aligned}
& =-18 \cdot \int_{1}^{5}\left(-\frac{1}{6 t}+\frac{1}{6(t-6)}\right) d t=3 \cdot \int_{1}^{5}\left(\frac{1}{t}-\frac{1}{t-6}\right) d t= \\
& =\left.3 \cdot(\ln |t|-\ln |t-6|)\right|_{1} ^{5}=3 \cdot(\ln |5|-\ln |5-6|)-3 \cdot(\ln |1|-\ln |1-6|)= \\
& =3 \cdot(\ln 5-\ln 1)-3 \cdot(\ln 1-\ln 5)=3 \cdot \ln 5+3 \cdot \ln 5=6 \ln 5
\end{aligned}
$$

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%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D1%8B_%D0%97%D0%B0%D0%B4%D0%B0%D1%87%D0%B0_9-25(2) »

Categories: Kuznetsov's Problem Book Integrals Problem 9 | Integrals

- Last modified: 15:37, 12 June 2009.
- Content is available under CC-BY-SA 3.0.

Created by GeeTeatoo

## Problem Kuznetsov Integrals 9-26

## Material from PlusPi
## Problem Statement

Calculate the definite integral:

$$
\int_{\pi / 4}^{\arccos (1 / \sqrt{26})} \frac{36 d x}{(6-\operatorname{tg} x) \sin 2 x}
$$