## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{\frac{\pi}{2}} \frac{\cos x \, dx}{2+\cos x}
$$
## 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:

$$
\begin{aligned}
& \int_{0}^{\frac{\pi}{2}} \frac{\cos x d x}{2+\cos x}=\int_{0}^{1} \frac{\frac{1-t^{2}}{1+t^{2}} \cdot \frac{2 d t}{1+t^{2}}}{2+\frac{1-t^{2}}{1+t^{2}}}=\int_{0}^{1} \frac{2\left(1-t^{2}\right) d t}{\left(1+t^{2}\right)^{2} \cdot \frac{2+2 t^{2}+1-t^{2}}{1+t^{2}}}= \\
& =2 \cdot \int_{0}^{1} \frac{1-t^{2}}{\left(1+t^{2}\right) \cdot\left(3+t^{2}\right)} d t=
\end{aligned}
$$

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

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

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

Subtract the first equation from the third:

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

Subtract the second equation from the fourth:

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

Then:

$$
\begin{aligned}
& =2 \cdot \int_{0}^{1}\left(\frac{1}{t^{2}+1}-\frac{2}{t^{2}+3}\right) d t=\left.2 \cdot\left(\operatorname{arctg} t-\frac{2}{\sqrt{3}} \cdot \operatorname{arctg} \frac{t}{\sqrt{3}}\right)\right|_{0} ^{1}= \\
& =2 \cdot\left(\operatorname{arctg} 1-\frac{2}{\sqrt{3}} \cdot \operatorname{arctg} \frac{1}{\sqrt{3}}\right)-2 \cdot\left(\operatorname{arctg} 0-\frac{2}{\sqrt{3}} \cdot \operatorname{arctg} \frac{0}{\sqrt{3}}\right)= \\
& =2 \cdot\left(\frac{\pi}{4}-\frac{2}{\sqrt{3}} \cdot \frac{\pi}{6}\right)-2 \cdot\left(0-\frac{2}{\sqrt{3}} \cdot 0\right)=2 \cdot\left(\frac{9 \pi}{36}-\frac{4 \sqrt{3} \pi}{36}\right)=\frac{(9-4 \sqrt{3}) \pi}{18}
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+8-2$ »

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

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## Problem Kuznetsov Integrals $8-3$

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