Discussion
The value of \(\int_{\pi/10}^{2\pi/5} \dfrac{cot^{3}x}{1 + cot^{3}x}\,dx\) is equal to
Answer: C
The integral is \(\frac{3\pi}{20}\).
The limits add to \(\frac{\pi}{10} + \frac{2\pi}{5} = \frac{\pi}{2}\), so use \(\int_{a}^{b}f(x)dx = \int_{a}^{b}f(a + b - x)dx\). Replacing \(x\) by \(\frac{\pi}{2} - x\) turns \(\cot x\) into \(\tan x\):
\[ \begin{aligned} I &= \int \frac{\cot^{3}x}{1 + \cot^{3}x}dx = \int \frac{\tan^{3}x}{1 + \tan^{3}x}dx \\ 2I &= \int_{\pi/10}^{2\pi/5} 1\,dx = \frac{3\pi}{10} \end{aligned} \]
(The two integrands add to 1, since \(\frac{\tan^{3}x}{1 + \tan^{3}x} = \frac{1}{\cot^{3}x + 1}\).) So \(I = \frac{3\pi}{20}\). \(\frac{3\pi}{10}\) is the length of the interval without halving; \(\frac{\pi}{4}\) is the familiar answer for limits 0 to \(\frac{\pi}{2}\).
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