## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{\frac{\pi}{4}} \frac{d x}{\cos x(1+\cos x)}
$$
## Solution

We will use the universal substitution:

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

From which:

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

Since $\operatorname{tg} \frac{x}{2}=\frac{1-\cos x}{\sin x}$, then

$$
x=\frac{\pi}{4} \Rightarrow t=\operatorname{tg} \frac{\left(\frac{\pi}{4}\right)}{2}=\frac{1-\cos \frac{\pi}{4}}{\sin \frac{\pi}{4}}=\frac{1-\frac{\sqrt{2}}{2}}{\left(\frac{\sqrt{2}}{2}\right)}=\frac{2-\sqrt{2}}{\sqrt{2}}
$$

Substitute:

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

$$
\begin{aligned}
& =-\frac{2-\sqrt{2}}{\sqrt{2}}+0-\ln \left|\frac{\frac{2-\sqrt{2}}{\sqrt{2}}-1}{\frac{2-\sqrt{2}}{\sqrt{2}}+1}\right|+\ln \left|\frac{0-1}{0+1}\right|= \\
& =\frac{\sqrt{2}-2}{\sqrt{2}}-\ln \left|\frac{2-\sqrt{2}-\sqrt{2}}{2-\sqrt{2}+\sqrt{2}}\right|+\ln 1=\frac{\sqrt{2}-2}{\sqrt{2}}-\ln (\sqrt{2}-1)
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_8-30$ »

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

Ukrainian Banner Network

- Last edited on this page: 20:30, 2 May 2009.
- Content is available under CC-BY-SA 3.0.

Created by GeeTeatoo

## Problem Kuznetsov Integrals 8-31

## Material from PlusPi