The ∫cos⁡(x−a)cos⁡(x−b)dx=

∫cos⁡(x−b+b−a)cos⁡(x−b)dx =∫cos⁡(x−b)cos⁡(b−a)−sin⁡(x−b)sin⁡(b−a)cos⁡(x−b)dx=∫cos⁡(b−a)dx−∫sin⁡(b−a)tan⁡(x−b)dxxcos⁡(b−a)−sin⁡(b−a)log⁡|sec⁡(x−b)|+c

The ∫cos⁡(x−a)cos⁡(x−b)dx=

∫cos⁡(x−b+b−a)cos⁡(x−b)dx =∫cos⁡(x−b)cos⁡(b−a)−sin⁡(x−b)sin⁡(b−a)cos⁡(x−b)dx=∫cos⁡(b−a)dx−∫sin⁡(b−a)tan⁡(x−b)dxxcos⁡(b−a)−sin⁡(b−a)log⁡|sec⁡(x−b)|+c

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The $\int \frac{\cos (x - a)}{\cos (x - b)} dx =$

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$xcos (b - a) - \sin (b - a) \log | \sec (x - b) | + c$

$xcos (b - a) + \sin (b - a) \log | \sec (x - b) | + c$

$xcos (b + a) - \sin (b + a) \log | \sec (x + b) | + c$

$xcos (b + a) - \sin (b + a) \log | \sec (x - b) | + c$

answer is A.

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Detailed Solution

$\int \frac{\cos (x - b + b - a)}{\cos (x - b)} dx \begin{array}{l} = \int \frac{\cos (x - b) \cos (b - a) - \sin (x - b) \sin (b - a)}{\cos (x - b)} dx \\ = \int \cos (b - a) dx - \int \sin (b - a) \tan (x - b) dx \\ xcos (b - a) - \sin (b - a) \log | \sec (x - b) | + c \end{array}$