So, we have:
$\int_{\pi / 6}^{\pi / 3} \cot ^{3} x d x = \int_{\pi / 6}^{\pi / 3} (\csc^2 x - 1) \cot x d x$
Now, let's use integration by parts. Let $u = \cot x$ and $dv = (\csc^2 x - 1) dx$. Then, $du = -\csc^2 x dx$ and $v = -\ln |\csc x + \cot x|$.
Using
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