int frac{sin (x-a)}{sin (x-b)} d x = A x + B | vedclass

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(A) To evaluate the integral $I = \int \frac{\sin (x-a)}{\sin (x-b)} dx$,we rewrite the numerator as $\sin((x-b) + (b-a))$.Using the identity $\sin(u+

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$(\cos (b-a), \sin (b-a))$

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(A) To evaluate the integral $I = \int \frac{\sin (x-a)}{\sin (x-b)} dx$,we rewrite the numerator as $\sin((x-b) + (b-a))$.Using the identity $\sin(u+v) = \sin u \cos v + \cos u \sin v$,we get:$I = \int \frac{\sin(x-b)\cos(b-a) + \cos(x-b)\sin(b-a)}{\sin(x-b)} dx$$I = \int \cos(b-a) dx + \int \sin(b-a) \cot(x-b) dx$$I = x \cos(b-a) + \sin(b-a) \log |\sin(x-b)| + C$Comparing this with $Ax + B \log |\sin(x-b)| + C$,we find $A = \cos(b-a)$ and $B = \sin(b-a)$.Thus,$(A, B) = (\cos(b-a), \sin(b-a))$.

$\int \frac{\sin (x-a)}{\sin (x-b)} d x = A x + B \log |\sin (x-b)| + C \Rightarrow (A, B) = $

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