I=∫1sin⁡(x−a)cos⁡(x−b)dx is equal to

I=∫1sin⁡(x−a)cos⁡(x−b)dxI=−1cos⁡(a−b)∫cos⁡{(x−a)−(x−b)}sin⁡(x−a)cos⁡(x−b)dxI=1cos⁡(a−b)∫{cot⁡(x−a)+tan⁡(x−b)}+dxI=1cos⁡(a−b){log⁡sin⁡(x−a)−log⁡cos⁡(x−b)}+CI=1cos⁡(a−b)log⁡sin⁡(x−a)cos⁡(x−b)+C

I=∫1sin⁡(x−a)cos⁡(x−b)dx is equal to

I=∫1sin⁡(x−a)cos⁡(x−b)dxI=−1cos⁡(a−b)∫cos⁡{(x−a)−(x−b)}sin⁡(x−a)cos⁡(x−b)dxI=1cos⁡(a−b)∫{cot⁡(x−a)+tan⁡(x−b)}+dxI=1cos⁡(a−b){log⁡sin⁡(x−a)−log⁡cos⁡(x−b)}+CI=1cos⁡(a−b)log⁡sin⁡(x−a)cos⁡(x−b)+C

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$I = \int \frac{1}{\sin (x - a) \cos (x - b)} d x$ is equal to

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$\frac{1}{\sin (a - b)} \log |\frac{\sin (x - a)}{\cos (x - b)}| + C$

$\frac{1}{\cos (a - b)} \log |\frac{\sin (x - a)}{| \cos (x - b) |}| + C$

$\frac{1}{\sin (a + b)} - \log |\frac{\sin (x - a)}{| \cos (x - b) |}| + C$

$\frac{1}{\cos (a + b)} \log |\frac{\sin (x - a)}{\cos (x - b)}| + C$

answer is B.

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Detailed Solution

$I = \int \frac{1}{\sin (x - a) \cos (x - b)} d x$

$\begin{array}{l} I = - \frac{1}{\cos (a - b)} \int \frac{\cos {(x - a) - (x - b)}}{\sin (x - a) \cos (x - b)} d x \\ I = \frac{1}{\cos (a - b)} \int {\cot (x - a) + \tan (x - b)} + d x \end{array}$

$\begin{aligned} I = \frac{1}{\cos (a - b)} {\log \sin (x - a) - \log \cos (x - b)} + C \\ I = \frac{1}{\cos (a - b)} \log |\frac{\sin (x - a)}{\cos (x - b)}| + C \end{aligned}$