## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{2 \operatorname{arctan} \frac{1}{2}} \frac{(1-\sin x) d x}{\cos x(1+\cos x)}
$$
## Solution

We will use the universal substitution:

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

From which:

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

Substitute:

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

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} \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D1} \% 8 \mathrm{~B}+8-20 »

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

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## Problem Kuznetsov Integrals $8-21$

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