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

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(B) To evaluate the integral $I = \int \frac{\sin x}{\sin (x - \alpha )} dx$,we rewrite the numerator as $\sin(x - \alpha + \alpha)$.Using the identit

Vedclass · Accepted Answer

$x \cos \alpha + \sin \alpha \log |\sin (x - \alpha )| + c$

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(B) To evaluate the integral $I = \int \frac{\sin x}{\sin (x - \alpha )} dx$,we rewrite the numerator as $\sin(x - \alpha + \alpha)$.Using the identity $\sin(A + B) = \sin A \cos B + \cos A \sin B$,we get:$I = \int \frac{\sin(x - \alpha) \cos \alpha + \cos(x - \alpha) \sin \alpha}{\sin(x - \alpha)} dx$$I = \int \left( \cos \alpha + \sin \alpha \cdot \cot(x - \alpha) \right) dx$$I = \int \cos \alpha \, dx + \sin \alpha \int \cot(x - \alpha) \, dx$Since $\int \cos \alpha \, dx = x \cos \alpha$ and $\int \cot(x - \alpha) \, dx = \log |\sin(x - \alpha)|$,we have:$I = x \cos \alpha + \sin \alpha \log |\sin(x - \alpha)| + c$.

$\int \frac{\sin x}{\sin (x - \alpha )} dx = $

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