## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{\arccos \sqrt{2 / 3}} \frac{\tan x + 2}{\sin^2 x + 2 \cos^2 x - 3} \, dx
$$
## Solution

$$
\begin{aligned}
& \int_{0}^{\arccos \sqrt{2 / 3}} \frac{\operatorname{tg} x+2}{\sin ^{2} x+2 \cos ^{2} x-3} d x=\int_{0}^{\arccos \sqrt{2 / 3}} \frac{\operatorname{tg} x+2}{\cos ^{2} x-2} d x= \\
& =\int_{0}^{\arccos \sqrt{2 / 3}} \frac{\operatorname{tg} x+2}{\frac{1}{2} \cdot \cos 2 x-\frac{3}{2}} d x=2 \cdot \int_{0}^{\arccos \sqrt{2 / 3}} \frac{\operatorname{tg} x+2}{\cos 2 x-3} d x=
\end{aligned}
$$

We will use the substitution:

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

From which:

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

Substitute:

$$
\begin{aligned}
& =2 \cdot \int_{0}^{\frac{1}{\sqrt{2}}} \frac{t+2}{\frac{1-t^{2}}{1+t^{2}}-3} \cdot \frac{d t}{1+t^{2}}=2 \cdot \int_{0}^{\frac{1}{\sqrt{2}}} \frac{t+2}{1-t^{2}-3-3 t^{2}} d t=2 \cdot \int_{0}^{\frac{1}{\sqrt{2}}} \frac{t+2}{-2-4 t^{2}} d t= \\
& =-\int_{0}^{\frac{1}{\sqrt{2}}} \frac{t+2}{1+2 t^{2}} d t=-\frac{1}{4} \cdot \int_{0}^{\frac{1}{\sqrt{2}}} \frac{4 t}{2 t^{2}+1} d t-\int_{0}^{\frac{1}{\sqrt{2}}} \frac{1}{t^{2}+\frac{1}{2}} d t=
\end{aligned}
$$

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

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 $\% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+9-6 \%$ Categories: Kuznetsov's Problem Book Integrals Problem 9 | Integrals

- Last edited: 21:38, 8 May 2009.
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## Problem Kuznetsov Integrals $9-7$

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