## Problem Statement

Calculate the definite integral:

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

Let's use the universal substitution:

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

From which:

$$
\begin{aligned}
& \sin x=\frac{2 t}{1+t^{2}}, \cos x=\frac{1-t^{2}}{1+t^{2}}, d x=\frac{2 d t}{1+t^{2}} \\
& x=\frac{\pi}{3} \Rightarrow t=\operatorname{tg} \frac{\left(\frac{\pi}{3}\right)}{2}=\operatorname{tg} \frac{\pi}{6}=\frac{1}{\sqrt{3}} \\
& x=\frac{\pi}{2} \Rightarrow t=\operatorname{tg} \frac{\left(\frac{\pi}{2}\right)}{2}=\operatorname{tg} \frac{\pi}{4}=1
\end{aligned}
$$

Substitute:

![](https://cdn.mathpix.com/cropped/2024_05_22_4691033a31e4eabc857cg-24.jpg?height=212&width=1357&top_left_y=1870&top_left_x=224)

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

$$
\begin{aligned}
& =\ln |t|_{\frac{1}{\sqrt{3}}}^{1}-\left.\frac{1}{2} \cdot \ln \left(1+t^{2}\right)\right|_{\frac{1}{\sqrt{3}}} ^{1}-\operatorname{arctg} 1+\operatorname{arctg} \frac{1}{\sqrt{3}}= \\
& =\ln |1|-\ln \left|\frac{1}{\sqrt{3}}\right|-\frac{1}{2} \cdot \ln \left(1+1^{2}\right)+\frac{1}{2} \cdot \ln \left(1+\left(\frac{1}{\sqrt{3}}\right)^{2}\right)-\frac{\pi}{4}+\frac{\pi}{6}= \\
& =\frac{1}{2} \cdot \ln \left(\frac{3}{2} \cdot \frac{4}{3}\right)-\frac{\pi}{12}=\frac{1}{2} \cdot \ln 2-\frac{\pi}{12}
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+8-11$ » Categories: Kuznetsov's Problem Book Integrals Problem $8 \mid$ Integrals

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## Problem Kuznetsov Integrals 8-12

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