If int frac{sin x}{sin (x-alpha)} dx = Ax + B | vedclass

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(A) Let $I = \int \frac{\sin x}{\sin (x-\alpha)} dx$.We can rewrite the numerator as $\sin x = \sin ((x-\alpha) + \alpha)$.Using the identity $\sin (A

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$\cos \alpha, \sin \alpha$

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(A) Let $I = \int \frac{\sin x}{\sin (x-\alpha)} dx$.We can rewrite the numerator as $\sin x = \sin ((x-\alpha) + \alpha)$.Using the identity $\sin (A+B) = \sin A \cos B + \cos A \sin B$,we get:$\sin ((x-\alpha) + \alpha) = \sin (x-\alpha) \cos \alpha + \cos (x-\alpha) \sin \alpha$.Substituting this into the integral:$I = \int \frac{\sin (x-\alpha) \cos \alpha + \cos (x-\alpha) \sin \alpha}{\sin (x-\alpha)} dx$.$I = \int \cos \alpha dx + \int \sin \alpha \frac{\cos (x-\alpha)}{\sin (x-\alpha)} dx$.$I = \cos \alpha \int dx + \sin \alpha \int \cot (x-\alpha) dx$.Integrating,we get $I = x \cos \alpha + \sin \alpha \log |\sin (x-\alpha)| + c$.Comparing this with the given form $Ax + B \log |\sin (x-\alpha)| + c$,we find $A = \cos \alpha$ and $B = \sin \alpha$.

If $\int \frac{\sin x}{\sin (x-\alpha)} dx = Ax + B \log |\sin (x-\alpha)| + c$,then the values of $A$ and $B$ are respectively (where $c$ is a constant of integration).

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