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

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

Vedclass · Accepted Answer

$\csc(a - b) \ln \left| \frac{\cos(x - a)}{\cos(x - b)} \right| + C$

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

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

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