## Problem Statement

Calculate the definite integral:

$$
\int_{-\arccos (1 / \sqrt{5})}^{0} \frac{11-3 \tan x}{\tan x+3} d x
$$
## Solution

Let's use the substitution:

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

From which we get:

$$
\begin{aligned}
& d x=\frac{d t}{1+t^{2}} \\
& x=-\arccos \frac{1}{\sqrt{5}} \Rightarrow t=\operatorname{tg}\left(-\arccos \frac{1}{\sqrt{5}}\right)=-\operatorname{tg}\left(\arccos \frac{1}{\sqrt{5}}\right)= \\
& =-\sqrt{\frac{1}{\cos ^{2}\left(\arccos \frac{1}{\sqrt{5}}\right)}-1}=-\sqrt{\frac{1}{\left(\frac{1}{\sqrt{5}}\right)^{2}}-1}=-\sqrt{5-1}=-\sqrt{4}=-2 \\
& x=0 \Rightarrow t=\operatorname{tg} 0=0
\end{aligned}
$$

Substitute:

$$
\int_{-\arccos (1 / \sqrt{5})}^{0} \frac{11-3 \operatorname{tg} x}{\operatorname{tg} x+3} d x=\int_{-2}^{0} \frac{11-3 t}{t+3} \cdot \frac{d t}{1+t^{2}}=\int_{-2}^{0} \frac{11-3 t}{(t+3)\left(1+t^{2}\right)} d t=
$$

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

$$
\begin{aligned}
& \frac{11-3 t}{(t+3)\left(1+t^{2}\right)}=\frac{A}{t+3}+\frac{B t+C}{1+t^{2}}= \\
& =\frac{A\left(1+t^{2}\right)+(B t+C)(t+3)}{(t+3)\left(1+t^{2}\right)}= \\
& =\frac{A+A t^{2}+B t^{2}+3 B t+C t+3 C}{(t+3)\left(1+t^{2}\right)}=
\end{aligned}
$$

$$
\begin{aligned}
& =\frac{(A+B) t^{2}+(3 B+C) t+(A+3 C)}{(t+3)\left(1+t^{2}\right)} \\
& \left\{\begin{array}{l}
A+B=0 \\
3 B+C=-3 \\
A+3 C=11
\end{array}\right. \\
& \left\{\begin{array}{l}
A=-B \\
C=-3-3 B \\
A+3 C=11
\end{array}\right. \\
& \left\{\begin{array}{l}
A=-B \\
C=-3-3 B \\
-B-9-9 B=11
\end{array}\right. \\
& \left\{\begin{array}{l}
A=-B \\
C=-3-3 B \\
-10 B=20
\end{array}\right. \\
& \left\{\begin{array}{l}
A=2 \\
C=3 \\
B=-2
\end{array}\right. \\
& \frac{11-3 t}{(t+3)\left(1+t^{2}\right)}=\frac{2}{t+3}-\frac{2 t-3}{1+t^{2}}
\end{aligned}
$$

We get:

$$
\begin{aligned}
& =\int_{-2}^{0}\left(\frac{2}{t+3}-\frac{2 t-3}{1+t^{2}}\right) d t=2 \int_{-2}^{0} \frac{1}{t+3} d t-\int_{-2}^{0} \frac{2 t}{1+t^{2}} d t+3 \int_{-2}^{0} \frac{1}{1+t^{2}} d t= \\
& =\left.2 \ln |t+3|\right|_{-2} ^{0}-\int_{-2}^{0} \frac{d\left(1+t^{2}\right)}{1+t^{2}} d t+\left.3 \operatorname{arctg} t\right|_{-2} ^{0}= \\
& =2 \ln |0+3|-2 \ln |-2+3|-\left.\ln \left(1+t^{2}\right)\right|_{-2} ^{0}+3 \operatorname{arctg} 0-3 \operatorname{arctg}(-2)= \\
& =2 \ln 3-2 \ln 1-\ln \left(1+0^{2}\right)+\ln \left(1+(-2)^{2}\right)+0+3 \operatorname{arctg} 2= \\
& =2 \ln 3-0-\ln 1+\ln 5+3 \operatorname{arctg} 2=2 \ln 3+\ln 5+3 \operatorname{arctg} 2=\ln 45+3 \operatorname{arctg} 2
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+9-23$ » Categories: Kuznetsov's Integral Problems Problem 9 | Integrals

- Last edited: 17:14, 13 May 2009.
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