## Problem Statement

Calculate the definite integral:

$$
\int_{\frac{\pi}{2}}^{2 \operatorname{arctan} 2} \frac{d x}{\sin x(1+\sin x)}
$$
## Solution

## Method 1

We will use the universal substitution:

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

From which:

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

Substitute:

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

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

$$
\begin{aligned}
& \frac{1+t^{2}}{t(t+1)^{2}}=\frac{A}{t}+\frac{B_{1}}{t-1}+\frac{B_{2}}{(t-1)^{2}}=\frac{A(t+1)^{2}+B_{1} t(t+1)+B_{2} t}{t(t+1)^{2}}= \\
& =\frac{A\left(t^{2}+2 t+1\right)+B_{1}\left(t^{2}+t\right)+B_{2} t}{t(t+1)^{2}}=\frac{\left(A+B_{1}\right) t^{2}+\left(2 A+B_{1}+B_{2}\right) t+A}{t(t+1)^{2}} \\
& \left\{\begin{array}{l}
A+B_{1}=1 \\
2 A+B_{1}+B_{2}=0 \\
A=1
\end{array}\right. \\
& \left\{\begin{array}{l}
B_{1}=0 \\
B_{1}+B_{2}=-2 \\
A=1
\end{array}\right. \\
& \left\{\begin{array}{l}
B_{1}=0 \\
B_{2}=-2 \\
A=1
\end{array}\right. \\
& \frac{1+t^{2}}{t(t+1)^{2}}=\frac{1}{t}-\frac{2}{(t+1)^{2}} \\
& =\int_{1}^{2}\left(\frac{1}{t}-\frac{2}{(t+1)^{2}}\right) d t=\left.\left(\ln |t|+\frac{2}{t+1}\right)\right|_{1} ^{2}=\left(\ln |2|+\frac{2}{2+1}\right)-\left(\ln |1|+\frac{2}{1+1}\right)= \\
& =\ln 2+\frac{2}{3}-0-1=\ln 2-\frac{1}{3}
\end{aligned}
$$

## Method 2

We will use the universal substitution:

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

From which:

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

Substitute:

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

$$
\begin{aligned}
& =\int_{1}^{2}\left(\frac{1}{t}-\frac{2}{(t+1)^{2}}\right) d t=\left.\left(\ln |t|+\frac{2}{t+1}\right)\right|_{1} ^{2}=\left(\ln |2|+\frac{2}{2+1}\right)-\left(\ln |1|+\frac{2}{1+1}\right)= \\
& =\ln 2+\frac{2}{3}-0-1=\ln 2-\frac{1}{3}
\end{aligned}
$$

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\%D1\%82 $\% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_8-27$ » Categories: Kuznetsov's Integral Problems 8 | Integrals

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## Problem Kuznetsov Integrals 8-28

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