If int frac{sin x}{sin (x-alpha)} dx = px - q | vedclass

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Question

(A) To solve the integral $I = \int \frac{\sin x}{\sin (x-\alpha)} dx$,we substitute $u = x - \alpha$,so $x = u + \alpha$ and $dx = du$. Substituting

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$-\frac{1}{2} \sin 2\alpha$

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(A) To solve the integral $I = \int \frac{\sin x}{\sin (x-\alpha)} dx$,we substitute $u = x - \alpha$,so $x = u + \alpha$ and $dx = du$. Substituting these into the integral,we get: $I = \int \frac{\sin (u + \alpha)}{\sin u} du$ Using the identity $\sin (A+B) = \sin A \cos B + \cos A \sin B$: $I = \int \frac{\sin u \cos \alpha + \cos u \sin \alpha}{\sin u} du$ $I = \int (\cos \alpha + \cot u \sin \alpha) du$ $I = \cos \alpha \int du + \sin \alpha \int \cot u du$ $I = u \cos \alpha + \sin \alpha \log |\sin u| + c$ Substituting back $u = x - \alpha$: $I = (x - \alpha) \cos \alpha + \sin \alpha \log |\sin (x - \alpha)| + c$ $I = x \cos \alpha - \alpha \cos \alpha + \sin \alpha \log |\sin (x - \alpha)| + c$ Comparing this with the given form $px - q \log |\sin (x - \alpha)| + c$: $p = \cos \alpha$ $-q = \sin \alpha \implies q = -\sin \alpha$ Therefore,$pq = (\cos \alpha)(-\sin \alpha) = -\sin \alpha \cos \alpha = -\frac{1}{2} (2 \sin \alpha \cos \alpha) = -\frac{1}{2} \sin 2\alpha$.

If $\int \frac{\sin x}{\sin (x-\alpha)} dx = px - q \log |\sin (x-\alpha)| + c$,then $pq =$ . . . . . . .

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