Chapter 5 - The Integral - 5.4 The Fundamental Theorem of Calculus, Part I - Exercises - Page 258: 31

$$\frac{-1+\sqrt 2}{5}.$$

Since $(\csc 5x )'=-5\csc 5x\cot 5x $, we have $$\int_{\pi/20}^{\pi / 10} \csc 5x\cot 5x d x=-\frac{1}{5}\csc 5x |_{\pi/20}^{\pi / 10}\\ =-\frac{1}{5}(\csc \frac{\pi}{2}-\csc \frac{\pi}{4})=\frac{-1+\sqrt 2}{5}.$$