$\int \frac{cos (A - B)}{sin (x - A) cos (x - B)} d x =$

log | \frac{cos (x - B)}{sin (x - A)} | + c

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log | \frac{sin (x - A)}{cos (x - B)} | + c

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log | \frac{sin (x - B)}{sin (x - A)} | + c

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log | \frac{sin (x - A)}{sin (x - B)} | + c

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Solution

The correct option is B $log | \frac{sin (x - A)}{cos (x - B)} | + c$
$\int \frac{cos (A - B)}{sin (x - A) cos (x - B)} d x$

$\int \frac{cos [(x - B) - (x - A)]}{sin (x - a) cos (x - B)} d x$

$= \int \frac{cos (x - B) cos (x - A) + sin (x - B) sin (x - A)}{sin (x - A) cos (x - B)}$

$= \int cot (x - A) d x + \int t a n (x - B) d x$

$= l o g | sin (x - A |) + l o g | sec (x - B) | d x$

$= l o g | sin (x - A) sec (x - B) | + c$

$= l o g ∣ ∣ ∣ \frac{sin (x - A)}{cos (x - B)} ∣ ∣ ∣ + c$

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