If I = int frac{dx}{sin(x-a) sin(x-b)},then I | vedclass

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(D) To evaluate $I = \int \frac{dx}{\sin(x-a) \sin(x-b)}$,multiply and divide by $\sin(b-a)$:$I = \frac{1}{\sin(b-a)} \int \frac{\sin((x-a) - (x-b))}{

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$\frac{1}{\sin(b-a)} \log \left|\frac{\sin(x-b)}{\sin(x-a)}\right| + c$,where $c$ is a constant of integration.

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(D) To evaluate $I = \int \frac{dx}{\sin(x-a) \sin(x-b)}$,multiply and divide by $\sin(b-a)$:$I = \frac{1}{\sin(b-a)} \int \frac{\sin((x-a) - (x-b))}{\sin(x-a) \sin(x-b)} dx$Using the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$:$I = \frac{1}{\sin(b-a)} \int \frac{\sin(x-a) \cos(x-b) - \cos(x-a) \sin(x-b)}{\sin(x-a) \sin(x-b)} dx$$I = \frac{1}{\sin(b-a)} \int \left( \frac{\cos(x-b)}{\sin(x-b)} - \frac{\cos(x-a)}{\sin(x-a)} \right) dx$$I = \frac{1}{\sin(b-a)} \int (\cot(x-b) - \cot(x-a)) dx$Integrating,we get:$I = \frac{1}{\sin(b-a)} [\log|\sin(x-b)| - \log|\sin(x-a)|] + c$$I = \frac{1}{\sin(b-a)} \log \left| \frac{\sin(x-b)}{\sin(x-a)} \right| + c$

If $I = \int \frac{dx}{\sin(x-a) \sin(x-b)}$,then $I$ is given by

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