## 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:

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

$$
=\frac{\sqrt{3}}{2}+\frac{1}{2} \cdot \ln 1-\frac{1}{2} \cdot \ln 4=\frac{\sqrt{3}}{2}-\ln 2
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_8-24 »$ Categories: Kuznetsov's Problem Book Integrals Problem 8 | Integrals

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## Problem Kuznetsov Integrals $8-25$

## Material from PlusPi
## Problem Statement

Calculate the definite integral:

$$
\int_{-\frac{2 \pi}{3}}^{0} \frac{\cos ^{2} x d x}{(1+\cos x-\sin x)^{2}}
$$