If∫1sin⁡(x−a)sin⁡(x−b)dx =A logsin(x-a)sin(x-b)+c

∫1sin⁡(x−a)sin⁡(x−b)dx=1sin⁡(a−b)∫sin⁡{(x−b)−(x−a)}sin⁡(x−a)sin⁡(x−b)dx =1sin⁡(a−b)∫sin⁡(x−b)cos⁡(x−a)−cos⁡(x−b)sin⁡(x−a)sin⁡(x−a)sin⁡(x−b)dx=1sin⁡(a−b)∫{cot⁡(x−a)−cot⁡(x−b)}dx=1sin⁡(a−b){log⁡|sin⁡(x−a)|−log⁡|sin⁡(x−b)|}+c=cosec⁡(a−b)log⁡sin⁡(x−a)sin⁡(x−b)+c

If∫1sin⁡(x−a)sin⁡(x−b)dx =A logsin(x-a)sin(x-b)+c

∫1sin⁡(x−a)sin⁡(x−b)dx=1sin⁡(a−b)∫sin⁡{(x−b)−(x−a)}sin⁡(x−a)sin⁡(x−b)dx =1sin⁡(a−b)∫sin⁡(x−b)cos⁡(x−a)−cos⁡(x−b)sin⁡(x−a)sin⁡(x−a)sin⁡(x−b)dx=1sin⁡(a−b)∫{cot⁡(x−a)−cot⁡(x−b)}dx=1sin⁡(a−b){log⁡|sin⁡(x−a)|−log⁡|sin⁡(x−b)|}+c=cosec⁡(a−b)log⁡sin⁡(x−a)sin⁡(x−b)+c

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$If \int \frac{1}{\sin (x - a) \sin (x - b)} d x = A \log |\frac{\sin (x - a)}{\sin (x - b)}| + c$

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$\sin (a - b)$

Sin(b-a)

cosec(b-a)

cosec(a-b)

answer is D.

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Detailed Solution

$\begin{array}{l} \int \frac{1}{\sin (x - a) \sin (x - b)} d x \\ = \frac{1}{\sin (a - b)} \int \frac{\sin {(x - b) - (x - a)}}{\sin (x - a) \sin (x - b)} d x \\ = \frac{1}{\sin (a - b)} \int \frac{\sin (x - b) \cos (x - a) - \cos (x - b) \sin (x - a)}{\sin (x - a) \sin (x - b)} d x \\ = \frac{1}{\sin (a - b)} \int {\cot (x - a) - \cot (x - b)} d x \\ = \frac{1}{\sin (a - b)} {\log | \sin (x - a) | - \log | \sin (x - b) |} + c \\ = cosec (a - b) \log |\frac{\sin (x - a)}{\sin (x - b)}| + c \end{array}$