## Problem Statement

Calculate the definite integral:

$$
\int_{\pi / 4}^{\operatorname{arctg} 3} \frac{d x}{(3 \operatorname{tg} x+5) \sin 2 x}
$$
## Solution

Let's use the substitution:

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

From which we get:

$$
\begin{aligned}
& \sin 2 x=\frac{2 t}{1+t^{2}}, d x=\frac{d t}{1+t^{2}} \\
& x=\frac{\pi}{4} \Rightarrow t=\operatorname{tg} \frac{\pi}{4}=1 \\
& x=\operatorname{arctg} 3 \Rightarrow t=\operatorname{tg}(\operatorname{arctg} 3)=3
\end{aligned}
$$

Substitute:

$$
\int_{\pi / 4}^{\operatorname{arctg} 3} \frac{d x}{(3 \operatorname{tg} x+5) \sin 2 x}=\int_{1}^{3} \frac{\frac{d t}{1+t^{2}}}{(3 t+5) \cdot \frac{2 t}{1+t^{2}}}=\frac{1}{2} \cdot \int_{1}^{3} \frac{d t}{t(3 t+5)}=
$$

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

$$
\begin{aligned}
& \frac{1}{t(3 t+5)}=\frac{A}{t}+\frac{B}{3 t+5}=\frac{A(3 t+5)+B t}{t(3 t+5)}=\frac{(3 A+B) t+5 A}{t(3 t+5)} \\
& \left\{\begin{array}{l}
3 A+B=0 \\
5 A=1
\end{array}\right. \\
& \left\{\begin{array}{l}
B=-\frac{3}{5} \\
A=\frac{1}{5}
\end{array}\right. \\
& \frac{1}{t(3 t+5)}=\frac{1}{5 t}-\frac{3}{5(3 t+5)}
\end{aligned}
$$

Then:

$$
\begin{aligned}
& =\frac{1}{2} \cdot \int_{1}^{3}\left(\frac{1}{5 t}-\frac{3}{5(3 t+5)}\right) d t=\frac{1}{10} \cdot\left(\int_{1}^{3} \frac{d t}{t}-\int_{1}^{3} \frac{d(3 t+5)}{3 t+5}\right)= \\
& =\left.\frac{1}{10} \cdot(\ln |t|-\ln |3 t+5|)\right|_{1} ^{3}=\left.\frac{1}{10} \cdot\left(\ln \left|\frac{t}{3 t+5}\right|\right)\right|_{1} ^{3}= \\
& =\frac{1}{10}\left(\ln \left|\frac{3}{3 \cdot 3+5}\right|-\ln \left|\frac{1}{3 \cdot 1+5}\right|\right)= \\
& =\frac{1}{10}\left(\ln \frac{3}{14}-\ln \frac{1}{8}\right)=\frac{1}{10} \ln \frac{3 \cdot 8}{14}=\frac{1}{10} \ln \frac{12}{7}
\end{aligned}
$$

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

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- Last edited: 08:53, May 3, 2009.
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## Problem Kuznetsov Integrals 9-2

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