## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{\arccos (1 / \sqrt{6})} \frac{3 \tan^{2} x-1}{\tan^{2} x+5}
$$
## Solution

We will use the substitution:

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

From which:

$$
\begin{aligned}
& d x=\frac{d t}{1+t^{2}} \\
& x=0 \Rightarrow t=\operatorname{tg} 0=0 \\
& x=\arccos \frac{1}{\sqrt{6}} \Rightarrow t=\operatorname{tg}\left(\arccos \frac{1}{\sqrt{6}}\right)=\sqrt{\frac{1}{\cos ^{2}\left(\arccos \frac{1}{\sqrt{6}}\right)}-1}= \\
& =\sqrt{\frac{1}{\left(\frac{1}{6}\right)}-1}=\sqrt{6-1}=\sqrt{5}
\end{aligned}
$$

Substitute:

$$
\int_{0}^{\arccos (1 / \sqrt{6})} \frac{3 \operatorname{tg}^{2} x-1}{\operatorname{tg}^{2} x+5}=\int_{0}^{\sqrt{5}} \frac{3 t^{2}-1}{t^{2}+5} \cdot \frac{d t}{1+t^{2}}=\int_{0}^{\sqrt{5}} \frac{3 t^{2}-1}{\left(t^{2}+5\right)\left(t^{2}+1\right)} d t=
$$

Decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:

$$
\begin{aligned}
& \frac{3 t^{2}-1}{\left(t^{2}+5\right)\left(t^{2}+1\right)}=\frac{A t+B}{5+t^{2}}+\frac{C t+D}{1+t^{2}}= \\
& =\frac{(A t+B) \cdot\left(1+t^{2}\right)+(C t+D) \cdot\left(5+t^{2}\right)}{\left(5+t^{2}\right)\left(1+t^{2}\right)}=
\end{aligned}
$$

$$
\begin{aligned}
& =\frac{A t+A t^{3}+B+B t^{2}+5 C t+C t^{3}+5 D+D t^{2}}{\left(5+t^{2}\right)\left(1+t^{2}\right)}= \\
& =\frac{(A+C) t^{3}+(B+D) t^{2}+(A+5 C) t+(B+5 D)}{\left(5+t^{2}\right)\left(1+t^{2}\right)} \\
& \left\{\begin{array}{l}
A+C=0 \\
B+D=3 \\
A+5 C=0 \\
B+5 D=-1
\end{array}\right. \\
& \left\{\begin{array}{l}
A=-C \\
B=3-D \\
-C+5 C=0 \\
3-D+5 D=-1
\end{array}\right. \\
& \left\{\begin{array}{l}
A=0 \\
B=3-D \\
C=0 \\
4 D=-4
\end{array}\right. \\
& \left\{\begin{array}{l}
A=0 \\
B=4 \\
C=0 \\
D=-1
\end{array}\right. \\
& \frac{3 t^{2}-1}{\left(t^{2}+5\right)\left(t^{2}+1\right)}=\frac{4}{5+t^{2}}-\frac{1}{1+t^{2}}
\end{aligned}
$$

We get:

$$
\begin{aligned}
& =\int_{0}^{\sqrt{5}}\left(\frac{4}{5+t^{2}}-\frac{1}{1+t^{2}}\right) d t=4 \cdot \int_{0}^{\sqrt{5}} \frac{1}{5+t^{2}} d t-\int_{0}^{\sqrt{5}} \frac{1}{1+t^{2}} d t= \\
& =\left.4 \cdot \frac{1}{\sqrt{5}} \cdot \operatorname{arctg} \frac{t}{\sqrt{5}}\right|_{0} ^{\sqrt{5}}-\left.\operatorname{arctg} t\right|_{0} ^{\sqrt{5}}= \\
& =\frac{4}{\sqrt{5}} \cdot \operatorname{arctg} \frac{\sqrt{5}}{\sqrt{5}}-\frac{4}{\sqrt{5}} \cdot \operatorname{arctg} \frac{0}{\sqrt{5}}-\operatorname{arctg} \sqrt{5}+\operatorname{arctg} 0= \\
& =\frac{4}{\sqrt{5}} \cdot \operatorname{arctg} 1-0-\operatorname{arctg} \sqrt{5}+0=\frac{4}{\sqrt{5}} \cdot \frac{\pi}{4}-\operatorname{arctg} \sqrt{5}= \\
& =\frac{\pi}{\sqrt{5}}-\operatorname{arctg} \sqrt{5}
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+9-31$ »

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

- Last edited: 17:34, 14 May 2009.
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