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$\int (\cot x \cot (x + α) + 1) d x =$

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$\cot α \log |\frac{\sin x}{\sin (x + α)}| + C$

$\log | \sin x \sin (x + α) | + x + C$

$\log | \sin x \cos (x + α) | + x + C$

$\tan α \log |\frac{\cos x}{\sin (x + α)}| + C$

answer is A.

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Detailed Solution

$\int (\cot x \cot (x + α) + 1) d x$

$= \int \frac{\cot x \cot (x + α) + 1}{\cot [(x + α) - x]} \times \cot α d x$

$= \int \frac{\cot x \cot (x + α) + 1}{\frac{\cot (x + α) \cot x + 1}{\cot x - \cot (x + α)}} \times \cot α d x$

$= \cot α \int [\cot x - \cot (x + α)] d x$

$= \cot α [\log | \sin x | - \log | \sin (x + α) |]$

$= \cot α l o g |\frac{\sin x}{\sin (x + α)}| + c$

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