int frac{dx}{sin(x-a) cos(x-b)} = | vedclass

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(B) To solve the integral $I = \int \frac{dx}{\sin(x-a) \cos(x-b)}$,multiply and divide by $\cos(a-b)$:$I = \frac{1}{\cos(a-b)} \int \frac{\cos((x-b)-

Vedclass · Accepted Answer

$\frac{1}{\cos(b-a)} \log \left|\frac{\sin(x-a)}{\cos(x-b)}\right| + C$

Vedclass · Answer

(B) To solve the integral $I = \int \frac{dx}{\sin(x-a) \cos(x-b)}$,multiply and divide by $\cos(a-b)$:$I = \frac{1}{\cos(a-b)} \int \frac{\cos((x-b)-(x-a))}{\sin(x-a) \cos(x-b)} dx$Using the identity $\cos(A-B) = \cos A \cos B + \sin A \sin B$,we get:$I = \frac{1}{\cos(a-b)} \int \frac{\cos(x-b)\cos(x-a) + \sin(x-b)\sin(x-a)}{\sin(x-a) \cos(x-b)} dx$$I = \frac{1}{\cos(a-b)} \int \left( \frac{\cos(x-b)\cos(x-a)}{\sin(x-a) \cos(x-b)} + \frac{\sin(x-b)\sin(x-a)}{\sin(x-a) \cos(x-b)} ight) dx$$I = \frac{1}{\cos(a-b)} \int (\cot(x-a) + 	an(x-b)) dx$Integrating term by term:$I = \frac{1}{\cos(a-b)} [\ln|\sin(x-a)| - \ln|\cos(x-b)|] + C$$I = \frac{1}{\cos(a-b)} \ln \left| \frac{\sin(x-a)}{\cos(x-b)} ight| + C$

$\int \frac{dx}{\sin(x-a) \cos(x-b)} = $

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