## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{\frac{\pi}{2}} \frac{\sin x d x}{5+3 \sin x}
$$
## 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:

$$
\begin{aligned}
& \int_{0}^{\frac{\pi}{2}} \frac{\sin x d x}{5+3 \sin x}=\int_{0}^{1} \frac{\frac{2 t}{1+t^{2}} \cdot \frac{2 d t}{1+t^{2}}}{5+3 \cdot \frac{2 t}{1+t^{2}}}=\int_{0}^{1} \frac{4 t \cdot d t}{\left(1+t^{2}\right)^{2} \cdot \frac{5+5 t^{2}+6 t}{1+t^{2}}}= \\
& =\int_{0}^{1} \frac{4 t \cdot d t}{\left(1+t^{2}\right) \cdot\left(5+6 t+5 t^{2}\right)}=
\end{aligned}
$$

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

$$
\begin{aligned}
& \frac{4 t}{\left(1+t^{2}\right) \cdot\left(5+6 t+5 t^{2}\right)}=\frac{A t+B}{t^{2}+1}+\frac{C t+D}{5 t^{2}+6 t+5}= \\
& =\frac{(A t+B)\left(5 t^{2}+6 t+5\right)+(C t+D)\left(t^{2}+1\right)}{\left(t^{2}+1\right)\left(5 t^{2}+6 t+5\right)}= \\
& =\frac{5 A t^{3}+6 A t^{2}+5 A t+5 B t^{2}+6 B t+5 B+C t^{3}+C t+D t^{2}+D}{\left(t^{2}+1\right)\left(5 t^{2}+6 t+5\right)}= \\
& =\frac{(5 A+C) t^{3}+(6 A+5 B+D) t^{2}+(5 A+6 B+C) t+(5 B+D)}{\left(t^{2}+1\right)\left(5 t^{2}+6 t+5\right)}
\end{aligned}
$$

$$
\left\{\begin{array}{l}
5 A+C=0 \\
6 A+5 B+D=0 \\
5 A+6 B+C=4 \\
5 B+D=0
\end{array}\right.
$$

Subtract the first equation from the third:

$$
\begin{aligned}
& \left\{\begin{array}{l}
5 A+C=0 \\
6 A+5 B+D=0 \\
6 B=4 \\
5 B+D=0
\end{array}\right. \\
& \left\{\begin{array}{l}
5 A+C=0 \\
6 A+D=-\frac{10}{3} \\
B=\frac{2}{3} \\
D=-\frac{10}{3}
\end{array}\right. \\
& \left\{\begin{array}{l}
5 A+C=0 \\
6 A=0 \\
B=\frac{2}{3} \\
D=-\frac{10}{3}
\end{array}\right. \\
& \left\{\begin{array}{l}
C=0 \\
A=0 \\
B=\frac{2}{3} \\
D=-\frac{10}{3}
\end{array}\right. \\
& \frac{4 t}{\left(1+t^{2}\right) \cdot\left(5+6 t+5 t^{2}\right)}=\frac{2}{3\left(t^{2}+1\right)}-\frac{10}{3\left(5 t^{2}+6 t+5\right)}=\frac{2}{3\left(t^{2}+1\right)}-\frac{2}{3\left(t^{2}+\frac{6}{5} t+1\right)}
\end{aligned}
$$

Then:

$$
\begin{aligned}
& =\int_{0}^{1}\left(\frac{2}{3\left(t^{2}+1\right)}-\frac{2}{3\left(t^{2}+\frac{6}{5} t+1\right)}\right) d t=\int_{0}^{1}\left(\frac{2}{3\left(t^{2}+1\right)}-\frac{2}{3\left(\left(t+\frac{3}{5}\right)^{2}+\frac{16}{25}\right)}\right) d t= \\
& =\frac{2}{3} \cdot \int_{0}^{1} \frac{d t}{t^{2}+1}-\frac{2}{3} \cdot \int_{0}^{1} \frac{1}{\left(t+\frac{3}{5}\right)^{2}+\frac{16}{25}} d t= \\
& =\left.\frac{2}{3} \cdot \operatorname{arctg} t\right|_{0} ^{1}-\left.\frac{2}{3} \cdot \frac{1}{\sqrt{\frac{16}{25}}} \operatorname{arctg} \frac{t+\frac{3}{5}}{\sqrt{\frac{16}{25}}}\right|_{0} ^{1}=\left.\frac{2}{3} \cdot \operatorname{arctg} t\right|_{0} ^{1}-\left.\frac{5}{6} \operatorname{arctg} \frac{5 t+3}{4}\right|_{0} ^{1}= \\
& =\frac{2}{3} \cdot \operatorname{arctg} 1-\frac{2}{3} \cdot \operatorname{arctg} 0-\frac{5}{6} \operatorname{arctg} \frac{5 \cdot 1+3}{4}+\frac{5}{6} \operatorname{arctg} \frac{5 \cdot 0+3}{4}= \\
& =\frac{2}{3} \cdot \frac{\pi}{4}-0-\frac{5}{6} \operatorname{arctg} 2+\frac{5}{6} \operatorname{arctg} \frac{3}{4}=\frac{\pi}{6}-\frac{5}{6} \operatorname{arctg} 2+\frac{5}{6} \operatorname{arctg} \frac{3}{4}= \\
& =\frac{\pi-5 \operatorname{arctg} 2+5 \operatorname{arctg} \frac{3}{4}}{6}
\end{aligned}
$$

Source — "http://pluspi.org/wiki/index.php/%D0%9A%D1%83%D0%B7%D0%BD%D0%B5%D1%86%D0%BE%D0%B2_%D0%BD%D1%82%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D1%8B_%D0%97%D0%B0%D0%B4%D0%B0%D1%87%D0%B0_8-31" Categories: Kuznetsov's Problem Book Integrals Problem 8 | Integrals

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- Last edited: 21:17, May 2, 2009.
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