## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{\operatorname{arctan} 3} \frac{4+\operatorname{tan} x}{2 \sin ^{2} x+18 \cos ^{2} x} d x
$$
## Solution

Let's use the substitution:

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

From which we get:

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

Substitute:

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

Source — “http://pluspi.org/wiki/index.php/%D0%97%D0%B0%D0%B4%D0%B0%D1%87%D0%B8_%D0%9A%D1%83%D0%B7%D0%BD%D0%B5%D1%86%D0%BE%D0%B2%D0%B0_%D0%98%D0%BD%D1%82%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D1%8B_9-13” Categories: Kuznetsov's Problem Book Integrals Problem 9 | Integrals

- Last modified: 08:05, 13 May 2009.
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Created by GeeTeatoo

## Problem Kuznetsov Integrals 9-14

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