## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{\pi / 4} \frac{4-7 \tan x}{2+3 \tan x} d x
$$
## Solution

Let's use the substitution:

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

From which we get:

$$
\begin{aligned}
& d x=\frac{d t}{1+t^{2}} \\
& x=0 \Rightarrow t=\operatorname{tg} 0=0 \\
& x=\frac{\pi}{4} \Rightarrow t=\operatorname{tg} \frac{\pi}{4}=1
\end{aligned}
$$

Substitute:

$$
\int_{0}^{\pi / 4} \frac{4-7 \operatorname{tg} x}{2+3 \operatorname{tg} x} d x=\int_{0}^{1} \frac{4-7 t}{2+3 t} \cdot \frac{d t}{1+t^{2}}=\int_{0}^{1} \frac{4-7 t}{(2+3 t)\left(1+t^{2}\right)} d t=
$$

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

$$
\begin{aligned}
& \frac{4-7 t}{(3 t+2)\left(1+t^{2}\right)}=\frac{A}{3 t+2}+\frac{B t+C}{1+t^{2}}=\frac{A\left(1+t^{2}\right)+(B t+C)(3 t+2)}{(3 t+2)\left(1+t^{2}\right)}= \\
& =\frac{A+A t^{2}+3 B t^{2}+2 B t+3 C t+2 C}{(3 t+2)\left(1+t^{2}\right)}=\frac{(A+3 B) t^{2}+(2 B+3 C) t+(A+2 C)}{(3 t+2)\left(1+t^{2}\right)} \\
& \left\{\begin{array}{l}
A+3 B=0 \\
2 B+3 C=-7 \\
A+2 C=4
\end{array}\right. \\
& \left\{\begin{array}{l}
A=-3 B \\
2 B+3 C=-7 \\
-3 B+2 C=4
\end{array}\right.
\end{aligned}
$$

Multiply the second equation by 3 and the third by 2:

$$
\left\{\begin{array}{l}
A=-3 B \\
6 B+9 C=-21 \\
-6 B+4 C=8
\end{array}\right.
$$

Add the third equation to the second:

$$
\begin{aligned}
& \left\{\begin{array}{l}
A=-3 B \\
13 C=-13 \\
-6 B+4 C=8
\end{array}\right. \\
& \left\{\begin{array}{l}
A=-3 B \\
C=-1 \\
-6 B-4=8
\end{array}\right. \\
& \left\{\begin{array}{l}
A=6 \\
C=-1 \\
B=-2
\end{array}\right. \\
& \frac{4-7 t}{(3 t+2)\left(1+t^{2}\right)}=\frac{6}{3 t+2}-\frac{2 t+1}{1+t^{2}}
\end{aligned}
$$

We get:

$$
\begin{aligned}
& =\int_{0}^{1}\left(\frac{6}{3 t+2}-\frac{2 t+1}{1+t^{2}}\right) d t=2 \cdot \int_{0}^{1} \frac{1}{t+\frac{2}{3}} d t-\int_{0}^{1} \frac{2 t}{1+t^{2}} d t-\int_{0}^{1} \frac{1}{1+t^{2}} d t= \\
& =\left.2 \cdot \ln \left|t+\frac{2}{3}\right|\right|_{0} ^{1}-\int_{0}^{1} \frac{d\left(t^{2}+1\right)}{1+t^{2}} d t-\left.\operatorname{arctg} t\right|_{0} ^{1}= \\
& =2 \cdot \ln \left|1+\frac{2}{3}\right|-2 \cdot \ln \left|0+\frac{2}{3}\right|-\left.\ln \left(t^{2}+1\right)\right|_{0} ^{1}-\operatorname{arctg} 1+\operatorname{arctg} 0= \\
& =2 \cdot \ln \frac{5}{3}-2 \cdot \ln \frac{2}{3}-\ln \left(1^{2}+1\right)+\ln \left(0^{2}+1\right)-\frac{\pi}{4}+0= \\
& =2 \cdot \ln \frac{5}{2}-\ln 2+\ln 1-\frac{\pi}{4}=\ln \frac{25}{4}-\ln 2-\frac{\pi}{4}=\ln \frac{25}{8}-\frac{\pi}{4}
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_ $\% \mathrm{D} 0 \% 9 \mathrm{~A} \% \mathrm{D} 1 \% 83 \% \mathrm{D} 0 \% \mathrm{~B} 7 \% \mathrm{D} 0 \% \mathrm{BD} \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 1 \% 86 \% \mathrm{D} 0 \% \mathrm{BE} \% \mathrm{D} 0 \% \mathrm{~B} 2 \mathrm{H} 0 \% \% 98 \% \mathrm{D} 0 \% \mathrm{BD}$ $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+9-26$ "

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

- Last modified: 18:04, 13 May 2009.
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