## Problem Statement

Calculate the definite integral:

$$
\int_{\pi}^{2 \pi} \frac{1-\cos x}{(x-\sin x)^{2}} d x
$$
## Solution

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

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_4-26 »$ Categories: Kuznetsov Problem Book Integrals Problem 4 | Integrals

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## Problem Kuznetsov Integrals 4-27

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