## Problem Statement

Calculate the definite integral:

$$
\int_{-\frac{2 \pi}{3}}^{0} \frac{\cos x \, dx}{1+\cos x-\sin x}
$$
## Solution

We will use the universal substitution:

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

From which:

$$
\begin{aligned}
& \sin x=\frac{2 t}{1+t^{2}}, \cos x=\frac{1-t^{2}}{1+t^{2}}, d x=\frac{2 d t}{1+t^{2}} \\
& x=-\frac{2 \pi}{3} \Rightarrow t=\operatorname{tg} \frac{\left(-\frac{2 \pi}{3}\right)}{2}=\operatorname{tg}\left(-\frac{\pi}{3}\right)=-\sqrt{3} \\
& x=0 \Rightarrow t=\operatorname{tg} \frac{0}{2}=\operatorname{tg} 0=0
\end{aligned}
$$

Substitute:

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

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+8-17$ »

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

Ukrainian Banner Network

- Last modified: 18:46, 1 May 2009.
- Content is available under CC-BY-SA 3.0.

Created by GeeTeatoo

## Problem Kuznetsov Integrals $8-18$

## Material from PlusPi