## Problem Statement

Calculate the definite integral:

$$
\int_{\pi / 4}^{\arccos (1 / \sqrt{3})} \frac{\tan x}{\sin ^{2} x-5 \cos ^{2} x+4} d x
$$
## Solution

Let's use the substitution:

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

From which we get:

$$
\begin{aligned}
& \sin ^{2} x=\frac{t^{2}}{1+t^{2}}, \cos ^{2} x=\frac{1}{1+t^{2}}, d x=\frac{d t}{1+t^{2}} \\
& x=\frac{\pi}{4} \Rightarrow t=\operatorname{tg} \frac{\pi}{4}=1 \\
& x=\arccos \frac{1}{\sqrt{3}} \Rightarrow t=\operatorname{tg}\left(\arccos \frac{1}{\sqrt{3}}\right)=\sqrt{\frac{1}{\cos ^{2}\left(\arccos \frac{1}{\sqrt{3}}\right)}-1}= \\
& =\sqrt{\frac{1}{\left(\frac{1}{3}\right)}-1}=\sqrt{3-1}=\sqrt{2}
\end{aligned}
$$

Substitute:

$$
\begin{aligned}
& \int_{\pi / 4}^{\arccos (1 / \sqrt{3})} \frac{\operatorname{tg} x}{\sin ^{2} x-5 \cos ^{2} x+4} d x=\int_{1}^{\sqrt{2}} \frac{t}{\frac{t^{2}}{1+t^{2}}-\frac{5}{1+t^{2}}+4} \cdot \frac{d t}{1+t^{2}}= \\
= & \int_{1}^{\sqrt{2}} \frac{t}{t^{2}-5+4+4 t^{2}} d t=\int_{1}^{\sqrt{2}} \frac{t}{5 t^{2}-1} d t=\frac{1}{10} \cdot \int_{1}^{\sqrt{2}} \frac{2 t}{t^{2}-\frac{1}{5}} d t= \\
= & \frac{1}{10} \cdot \int_{1}^{\sqrt{2}} \frac{d\left(t^{2}-\frac{1}{5}\right)}{t^{2}-\frac{1}{5}}=\left.\frac{1}{10} \cdot \ln \left|t^{2}-\frac{1}{5}\right|\right|_{1} ^{\sqrt{2}}=\frac{1}{10} \cdot \ln \left|(\sqrt{2})^{2}-\frac{1}{5}\right|-\frac{1}{10} \cdot \ln \left|1^{2}-\frac{1}{5}\right|= \\
= & \frac{1}{10} \cdot \ln \frac{9}{5}-\frac{1}{10} \cdot \ln \frac{4}{5}=\frac{1}{10} \cdot \ln \frac{9}{4}
\end{aligned}
$$

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%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D1%8B_%D0%97%D0%B0%D0%B4%D0%B0%D1%87%D0%B0_9-11»

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

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- Last modified: 07:18, 13 May 2009.
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## Problem Kuznetsov Integrals 9-12

## Material from PlusPi

## Contents

- 1 Problem Statement
- 2 Solution
- 2.1 Solution 1
- 2.2 Solution 2