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$\int \frac{d x}{\sin (x-a) \cos (x-b)}=$
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Verified Answer
The correct answer is:
$\frac{1}{\cos (b-a)} \log \left|\frac{\sin (x-a)}{\cos (x-b)}\right|+C$
$\int \frac{d x}{\sin (x-a) \cos (x-b)}$
$=\frac{1}{\cos (b-a)} \int \frac{\cos [(x-a)-(x-b)]}{\sin (x-a) \cos (x-b)}$
$=\frac{1}{\cos (b-a)}$
$\int\left[\frac{\cos (x-a) \cos (x-b)+\sin (x-a) \sin (x-b)}{\sin (x-a) \cos (x-b)}\right] d x$
$\begin{aligned} & =\frac{1}{\cos (b+a)} \int\left[\frac{\cos (x-a)}{\sin (x-a)}+\frac{\sin (x-b)}{\cos (x-b)}\right] d x \\ & =\frac{1}{\cos (b-a)} \quad[\log (\sin (x-a)-\log (\cos (x-b))]+C \\ & =\frac{1}{\cos (b-a)} \log \left|\frac{\sin (x-a)}{\operatorname{cós}(x-b)}\right|+C\end{aligned}$
$=\frac{1}{\cos (b-a)} \int \frac{\cos [(x-a)-(x-b)]}{\sin (x-a) \cos (x-b)}$
$=\frac{1}{\cos (b-a)}$
$\int\left[\frac{\cos (x-a) \cos (x-b)+\sin (x-a) \sin (x-b)}{\sin (x-a) \cos (x-b)}\right] d x$
$\begin{aligned} & =\frac{1}{\cos (b+a)} \int\left[\frac{\cos (x-a)}{\sin (x-a)}+\frac{\sin (x-b)}{\cos (x-b)}\right] d x \\ & =\frac{1}{\cos (b-a)} \quad[\log (\sin (x-a)-\log (\cos (x-b))]+C \\ & =\frac{1}{\cos (b-a)} \log \left|\frac{\sin (x-a)}{\operatorname{cós}(x-b)}\right|+C\end{aligned}$
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