## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{\operatorname{arctan} 2} \frac{12+\operatorname{tan} x}{3 \sin ^{2} x+12 \cos ^{2} x} d x
$$
## Solution

We will use the substitution:

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

From which:

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

Substitute:

$$
\begin{aligned}
& \int_{0}^{\operatorname{arctg} 2} \frac{12+\operatorname{tg} x}{3 \sin ^{2} x+12 \cos ^{2} x} d x=\int_{0}^{2} \frac{12+t}{\frac{3 t^{2}}{1+t^{2}}+\frac{12}{1+t^{2}}} \cdot \frac{d t}{1+t^{2}}=\int_{0}^{2} \frac{12+t}{3 t^{2}+12} d t= \\
& =\int_{0}^{2} \frac{12}{3 t^{2}+12} d t+\int_{0}^{2} \frac{t}{3 t^{2}+12} d t=4 \cdot \int_{0}^{2} \frac{1}{t^{2}+4} d t+\frac{1}{6} \cdot \int_{0}^{2} \frac{d\left(t^{2}+4\right)}{t^{2}+4}= \\
& =\left.4 \cdot \frac{1}{2} \cdot \operatorname{arctg} \frac{t}{2}\right|_{0} ^{2}+\left.\frac{1}{6} \cdot \ln \left(t^{2}+4\right)\right|_{0} ^{2}= \\
& =2 \operatorname{arctg} 1-2 \operatorname{arctg} 0+\frac{1}{6} \cdot \ln \left(2^{2}+4\right)-\frac{1}{6} \cdot \ln \left(0^{2}+4\right)= \\
& =2 \cdot \frac{\pi}{4}-0+\frac{\ln 8}{6}-\frac{\ln 4}{6}=\frac{\pi}{2}+\frac{\ln 2}{6}
\end{aligned}
$$

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%D1%82%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D1%8B_%D0%97%D0%B0%D0%B4%D0%B0%D1%87%D0%B0_%D0%98%D0%BD%D1%82%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D1%8B_9-14» Categories: Kuznetsov's Problem Book Integrals Problem 9 | Integrals

- Last edited: 08:10, 13 May 2009.
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Created by GeeTeatoo

## Problem Kuznetsov Integrals 9-15

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