## Problem Statement

Calculate the definite integral:

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

We will use the universal substitution:

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

From which:

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

Substitute:

$$
\begin{aligned}
& \int_{\frac{\pi}{2}}^{2 \operatorname{arctg} 2} \frac{d x}{\sin ^{2} x(1-\cos x)}=\int_{1}^{2} \frac{\frac{2 d t}{1+t^{2}}}{\left(\frac{2 t}{1+t^{2}}\right)^{2} \cdot\left(1-\frac{\left.1-t^{2}\right)}{1+t^{2}}\right)}= \\
& =\int_{1}^{2} \frac{\frac{2 d t}{1+t^{2}}}{\frac{4 t^{2}}{\left(1+t^{2}\right)^{2}} \cdot \frac{1+t^{2}-1+t^{2}}{1+t^{2}}}=\int_{1}^{2} \frac{d t}{\frac{2 t^{2}}{1+t^{2}} \cdot \frac{2 t^{2}}{1+t^{2}}}=\int_{1}^{2} \frac{\left(1+t^{2}\right)^{2} d t}{4 t^{4}}= \\
& =\int_{1}^{2} \frac{1+2 t^{2}+t^{4}}{4 t^{4}} d t=\int_{1}^{2}\left(\frac{1}{4 t^{4}}+\frac{1}{2 t^{2}}+\frac{1}{4}\right) d t= \\
& =-\frac{1}{12 t^{3}}-\frac{1}{2 t}+\left.\frac{t}{4}\right|_{1} ^{2}=\left(-\frac{1}{12 \cdot 2^{3}}-\frac{1}{2 \cdot 2}+\frac{2}{4}\right)-\left(-\frac{1}{12 \cdot 1^{3}}-\frac{1}{2 \cdot 1}+\frac{1}{4}\right)= \\
& =-\frac{1}{96}-\frac{1}{4}+\frac{1}{2}+\frac{1}{12}+\frac{1}{2}-\frac{1}{4}=\frac{55}{96}
\end{aligned}
$$

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## Problem Kuznetsov Integrals 8-2

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