## Condition of the problem

Calculate the definite integral:

$$
\int_{0}^{\operatorname{arctg} \frac{1}{3}} \frac{8+\operatorname{tg} x}{18 \sin ^{2} x+2 \cos ^{2} x} d x
$$
## Solution

$$
\int_{0}^{\operatorname{arctg} \frac{1}{3}} \frac{8+\operatorname{tg} x}{18 \sin ^{2} x+2 \cos ^{2} x} d x=
$$

Let's use the substitution:

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

From which we get:

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

Substitute:

$$
\begin{aligned}
& =\int_{0}^{\frac{1}{3}} \frac{8+t}{\frac{18 t^{2}}{1+t^{2}}+\frac{2}{1+t^{2}}} \cdot \frac{d t}{1+t^{2}}=\int_{0}^{\frac{1}{3}} \frac{8+t}{18 t^{2}+2} d t=\frac{1}{18} \cdot \int_{0}^{\frac{1}{3}} \frac{8+t}{t^{2}+\frac{1}{9}} d t= \\
& =\frac{4}{9} \cdot \int_{0}^{\frac{1}{3}} \frac{d t}{t^{2}+\frac{1}{9}}+\frac{1}{36} \cdot \int_{0}^{\frac{1}{3}} \frac{2 t}{t^{2}+\frac{1}{9}} d t=\left.\frac{4}{9} \cdot \frac{1}{\left(\frac{1}{3}\right)} \cdot \operatorname{arctg} \frac{t}{\left(\frac{1}{3}\right)}\right|_{0} ^{\frac{1}{3}}+\frac{1}{36} \cdot \int_{0}^{\frac{1}{3}} \frac{d\left(t^{2}+\frac{1}{9}\right)}{t^{2}+\frac{1}{9}}= \\
& =\left.\frac{4}{3} \cdot \operatorname{arctg} 3 t\right|_{0} ^{\frac{1}{3}}+\left.\frac{1}{36} \cdot \ln \left(t^{2}+\frac{1}{9}\right)\right|_{0} ^{\frac{1}{3}}=\frac{4}{3} \cdot \frac{\pi}{4}-0+\frac{1}{36} \cdot \ln \left(\frac{1}{9}+\frac{1}{9}\right)-\frac{1}{36} \cdot \ln \frac{1}{9}= \\
& =\frac{\pi}{3}+\frac{1}{36} \cdot\left(\ln \frac{2}{9}+\ln 9\right)=\frac{\pi}{3}+\frac{\ln 2}{36}
\end{aligned}
$$

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

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

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