## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{\frac{\pi}{2}} \frac{\sin x \, dx}{2+\sin x}
$$
## Solution

We will use the universal substitution:

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

From which:

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

Substitute:

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

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

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

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

Subtract the first equation from the third:

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

Then:

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

$$
=\frac{\pi}{2}-\frac{4}{\sqrt{3}} \cdot \frac{\pi}{3}+\frac{4}{\sqrt{3}} \cdot \frac{\pi}{6}=\frac{\pi}{2}-\frac{4 \pi}{3 \sqrt{3}}+\frac{2 \pi}{3 \sqrt{3}}=\frac{\pi}{2}-\frac{2 \pi}{3 \sqrt{3}}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_8-29$ » Categories: Kuznetsov's Problem Book Integrals Problem 8 | Integrals

- Last edited: 19:53, 2 May 2009.
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## Problem Kuznetsov Integrals 8-30

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