$\int \frac{sin x}{sin (x - a)} d x = A x + B log sin (x - a) + c$ , then value of $(A, B)$ is

(sin a, cos a)

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(- cos a, sin a)

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(- sin a, cos a)

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(cos a, sin a)

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Solution

The correct option is D $(cos a, sin a)$
$\int \frac{sin x}{sin (x - a)} d x$
$= \int \frac{sin (x - a + a)}{sin (x - a)} d x$
$= \int \frac{sin (x - a) cos a + cos (x - a) sin a}{sin (x - a)} d x$
$= \int (cos a + cot (x - a) sin a) d x$
$= \int cos a d x + \int cot (x - a) sin a d x$
$= x cos a d x + sin a \int cot (x - a) d x$
$= x cos a + sin a l n | sin (x - 1) | + c$
$= A x + B log | sin (x - a) | + c$
So, $A = cos a$
$B = sin a$
$\Rightarrow (A, B) = (cos a, sin a)$
Hence, the answer is $(cos a, sin a) .$

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