## Problem Statement

Calculate the definite integral:

$$
\int_{-\operatorname{arctan}(1 / 3)}^{0} \frac{3 \tan x + 1}{2 \sin 2x - 5 \cos 2x + 1} d x
$$
## Solution

Let's use the substitution:

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

From which we get:

$$
\begin{aligned}
& \sin 2 x=\frac{2 t}{1+t^{2}}, \cos 2 x=\frac{1-t^{2}}{1+t^{2}}, d x=\frac{d t}{1+t^{2}} \\
& x=-\operatorname{arctg} \frac{1}{3} \Rightarrow t=\operatorname{tg}\left(-\operatorname{arctg} \frac{1}{3}\right)=-\operatorname{tg}\left(\operatorname{arctg} \frac{1}{3}\right)=-\frac{1}{3} \\
& x=0 \Rightarrow t=\operatorname{tg} 0=0
\end{aligned}
$$

Substitute:

$$
\begin{aligned}
& \int_{-\operatorname{arctg}(1 / 3)}^{0} \frac{3 \operatorname{tg} x+1}{2 \sin 2 x-5 \cos 2 x+1} d x=\int_{-\frac{1}{3}}^{0} \frac{3 t+1}{2 \cdot \frac{2 t}{1+t^{2}}-5 \cdot \frac{1-t^{2}}{1+t^{2}}+1} \cdot \frac{d t}{1+t^{2}}= \\
= & \int_{-\frac{1}{3}}^{0} \frac{3 t+1}{4 t-5+5 t^{2}+1+t^{2}} d t=\int_{-\frac{1}{3}}^{0} \frac{3 t+1}{6 t^{2}+4 t-4} d t=\frac{1}{4} \cdot \int_{-\frac{1}{3}}^{0} \frac{6 t+2}{3 t^{2}+2 t-2} d t= \\
= & \frac{1}{4} \cdot \int_{-\frac{1}{3}}^{0} \frac{d\left(3 t^{2}+2 t-2\right)}{3 t^{2}+2 t-2} d t=\left.\frac{1}{4} \cdot \ln \left|3 t^{2}+2 t-2\right|\right|_{-\frac{1}{3}} ^{0}= \\
= & \frac{1}{4} \cdot \ln \left|3 \cdot 0^{2}+2 \cdot 0-2\right|-\frac{1}{4} \cdot \ln \left|3 \cdot\left(-\frac{1}{3}\right)^{2}+2\left(-\frac{1}{3}\right)-2\right|= \\
= & \frac{\ln 2}{4}-\frac{1}{4} \cdot \ln \left|\frac{1}{3}-\frac{2}{3}-2\right|=\frac{\ln 2}{4}-\frac{1}{4} \cdot \ln \frac{7}{3}=\frac{1}{4} \ln \frac{6}{7}
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_9-9$ »

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

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## Problem Kuznetsov Integrals 9-10

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