Discussion
\(\displaystyle \int (\cot 2x + \mathrm{cosec}\, 2x)\,dx =\)
Answer: E
The integral is \(\log_{e}|\sin x| + C\).
Combine the two terms with the double-angle formulas:
\[ \begin{aligned} \cot 2x + \mathrm{cosec}\,2x &= \frac{\cos 2x + 1}{\sin 2x} \\ &= \frac{2\cos^{2}x}{2\sin x\cos x} = \cot x \end{aligned} \]
\[ \int \cot x\,dx = \log_{e}|\sin x| + C \]
Integrating each term separately and then forgetting to simplify can suggest a factor of \(\frac{1}{2}\) or 2, but the combined integrand is exactly \(\cot x\).
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