## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{\operatorname{arctan}(2 / 3)} \frac{6+\operatorname{tan} x}{9 \sin ^{2} x+4 \cos ^{2} x} d x
$$
## Solution

Let's use the substitution:

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

From which we get:

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

Substitute:

$$
\begin{aligned}
& \int_{0}^{\operatorname{arctg}(2 / 3)} \frac{6+\operatorname{tg} x}{9 \sin ^{2} x+4 \cos ^{2} x} d x=\int_{0}^{\frac{2}{3}} \frac{6+t}{\frac{9 t^{2}}{1+t^{2}}+\frac{4}{1+t^{2}}} \cdot \frac{d t}{1+t^{2}}=\int_{0}^{\frac{2}{3}} \frac{6+t}{9 t^{2}+4} d t= \\
& =\int_{0}^{\frac{2}{3}} \frac{6}{9 t^{2}+4} d t+\int_{0}^{\frac{2}{3}} \frac{t}{9 t^{2}+4} d t=\frac{2}{3} \cdot \int_{0}^{\frac{2}{3}} \frac{1}{t^{2}+\frac{4}{9}} d t+\frac{1}{18} \cdot \int_{0}^{\frac{2}{3}} \frac{d\left(t^{2}+\frac{4}{9}\right)}{t^{2}+\frac{4}{9}}= \\
& =\left.\frac{2}{3} \cdot \frac{1}{\left(\frac{2}{3}\right)} \cdot \operatorname{arctg} \frac{t}{\left(\frac{2}{3}\right)}\right|_{0} ^{\frac{2}{3}}+\left.\frac{1}{18} \cdot \ln \left(t^{2}+\frac{4}{9}\right)\right|_{0} ^{\frac{2}{3}}= \\
& =\operatorname{arctg} 1-\operatorname{arctg} 0+\frac{1}{18} \cdot \ln \left(\left(\frac{2}{3}\right)^{2}+\frac{4}{9}\right)-\frac{1}{18} \cdot \ln \left(0^{2}+\frac{4}{9}\right)= \\
& =\frac{\pi}{4}-0+\frac{1}{18} \ln \frac{8}{9}-\frac{1}{18} \ln \frac{4}{9}=\frac{\pi}{4}+\frac{\ln 2}{18}
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_9-15$ »

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

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