## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{\frac{\pi}{2}} \frac{\cos x d x}{5+4 \cos x}
$$
## Solution

We will use the universal substitution:

$$
t=\operatorname{tg} \frac{x}{2}
$$

From which:

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

Substitute:

$$
\begin{aligned}
& \int_{0}^{\frac{\pi}{2}} \frac{\cos x d x}{5+4 \cos x}=\int_{0}^{1} \frac{\frac{1-t^{2}}{1+t^{2}} \cdot \frac{2 d t}{1+t^{2}}}{5+4 \cdot \frac{1-t^{2}}{1+t^{2}}}=\int_{0}^{1} \frac{2\left(1-t^{2}\right) d t}{\left(1+t^{2}\right)^{2} \cdot \frac{5+5 t^{2}+4-4 t^{2}}{1+t^{2}}}= \\
& =2 \cdot \int_{0}^{1} \frac{1-t^{2}}{\left(1+t^{2}\right)\left(9+t^{2}\right)} d t=
\end{aligned}
$$

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

$$
\begin{aligned}
& \frac{1-t^{2}}{\left(1+t^{2}\right) \cdot\left(9+t^{2}\right)}=\frac{A t+B}{1+t^{2}}+\frac{C t+D}{9+t^{2}}= \\
& =\frac{(A t+B)\left(9+t^{2}\right)+(C t+D)\left(1+t^{2}\right)}{\left(1+t^{2}\right)\left(9+t^{2}\right)}= \\
& =\frac{A t^{3}+9 A t+B t^{2}+9 B+C t^{3}+C t+D t^{2}+D}{\left(1+t^{2}\right)\left(9+t^{2}\right)}=
\end{aligned}
$$

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

Subtract the first equation from the third:

$$
\begin{aligned}
& \left\{\begin{array}{l}
A+C=0 \\
B+D=-1 \\
8 A=0
\end{array}\right. \\
& \left\{\begin{array}{l}
C=0 \\
B+D=-1 \\
A=0 \\
9 B+D=1
\end{array}\right. \\
& \left\{\begin{array}{l}
C=0 \\
B+D=D
\end{array}\right.
\end{aligned}
$$

Subtract the second equation from the fourth:

$$
\begin{aligned}
& \left\{\begin{array}{l}
C=0 \\
B+D=-1 \\
A=0 \\
8 B=2
\end{array}\right. \\
& \left\{\begin{array}{l}
C=0 \\
D=-\frac{5}{4} \\
A=0 \\
B=\frac{1}{4}
\end{array}\right. \\
& \frac{1-t^{2}}{\left(1+t^{2}\right)\left(9+t^{2}\right)}=\frac{1}{4\left(1+t^{2}\right)}-\frac{5}{4\left(9+t^{2}\right)}
\end{aligned}
$$

Then:

$$
\begin{aligned}
& =2 \cdot \int_{0}^{1}\left(\frac{1}{4\left(1+t^{2}\right)}-\frac{5}{4\left(9+t^{2}\right)}\right) d t=\left.2 \cdot\left(\frac{1}{4} \cdot \operatorname{arctg} t-\frac{5}{4} \cdot \frac{1}{3} \cdot \operatorname{arctg} \frac{t}{3}\right)\right|_{0} ^{1}= \\
& =2 \cdot\left(\frac{1}{4} \cdot \operatorname{arctg} 1-\frac{5}{12} \cdot \operatorname{arctg} \frac{1}{3}\right)-2 \cdot\left(\frac{1}{4} \cdot \operatorname{arctg} 0-\frac{5}{12} \cdot \operatorname{arctg} \frac{0}{3}\right)= \\
& =2 \cdot\left(\frac{1}{4} \cdot \frac{\pi}{4}-\frac{5}{12} \cdot \operatorname{arctg} \frac{1}{3}\right)-2 \cdot\left(\frac{1}{4} \cdot 0-\frac{5}{12} \cdot 0\right)=\frac{\pi}{8}-\frac{5}{6} \cdot \operatorname{arctg} \frac{1}{3}
\end{aligned}
$$

Kuznetsov Integral Problem 8

Categories: Kuznetsov's Problem Book Integrals Problem 8 | Integrals

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- Last edited: 08:12, April 29, 2009.
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## Problem Kuznetsov Integrals 8-10

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