∫1sin⁡(x−a)sin⁡(x−b)dx=

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1sin⁡(a−b)log⁡sin⁡(x−a)sin⁡(x−b)+C

−1sin⁡(a−b)log⁡sin⁡(x−a)sin⁡(x−b)+C

log⁡sin⁡(x−a)sin⁡(x−b)+C

answer is A.

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Detailed Solution

We have, ∫1sin⁡(x−a)sin⁡(x−b)dx=1sin⁡(b−a)∫sin⁡{(x−a)−(x−b)}sin⁡(x−a)sin⁡(x−b)dx=1sin⁡(b−a)∫{cot⁡(x−b)−cot⁡(x−a)}dx=1sin⁡(b−a){log⁡sin⁡(x−b)−log⁡sin⁡(x−a)}+C=1sin⁡(b−a)log⁡sin⁡(x−b)sin⁡(x−a)+C=1sin⁡(a−b)log⁡sin⁡(x−a)sin⁡(x−b)+C

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