## Problem Statement

Calculate the definite integral:

$$
\int_{\pi}^{2 \pi} \frac{x+\cos x}{x^{2}+2 \sin x} d x
$$
## Solution

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

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+4-5$ "

Categories: Kuznetsov's Problem Book Integrals Problem 4 | Integrals

- Last edited: March 3, 2009, at 14:58.
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## Problem Kuznetsov Integrals 4-6

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