Question

\[\int\frac{1}{\text{ cos }\left( x - a \right) \text{ cos }\left( x - b \right)} \text{ dx }\]

Answer 1

\[\int\frac{1}{\text{ cos } \left( x - a \right) \cdot \text{ cos} \left( x - b \right)}dx\]
\[ = \frac{1}{\text{ sin }\left( a - b \right)}\int\frac{\text{ sin }\left( a - b \right)}{\text{ cos}\left( x - a \right) \cdot \text{ cos}\left( x - b \right)} dx\]
\[ = \frac{1}{\text{ sin }\left( a - b \right)}\int\frac{\text{ sin }\left[ \left( x - b \right) - \left( x - a \right) \right]}{\text{ cos}\left( x - a \right) \cdot \text{ cos}\left( x - b \right)} dx\]
\[ = \frac{1}{\text{ sin }\left( a - b \right)}\int\frac{\text{ sin }\left( x - b \right) \cdot \text{ cos}\left( x - a \right) - \text{ cos}\left( x - b \right) \cdot \text{ sin }\left( x - a \right)}{\text{ cos}\left( x - a \right) \cdot \text{ cos}\left( x - b \right)}\]
\[ = \frac{1}{\text{ sin }\left( a - b \right)}\int\left[ \frac{\text{ sin }\left( x - b \right) \cdot \text{ cos}\left( x - a \right)}{\text{ cos}\left( x - a \right) \cdot \text{ cos}\left( x - b \right)} - \frac{\text{ cos}\left( x - b \right) \cdot \text{ sin }\left( x - a \right)}{\text{ cos}\left( x - a \right) \cdot \text{ cos}\left( x - b \right)} \right] dx\]
\[ = \frac{1}{\text{ sin }\left( a - b \right)}\int\left[ \text{ tan }\left( x - b \right) - \text{ tan }\left( x - a \right) \right] dx\]
\[ = \frac{1}{\text{ sin }\left( a - b \right)}\int\text{ tan }\left( x - b \right) dx - \int\text{ tan } \left( x - a \right) dx\]
\[ = \frac{1}{\text{ sin }\left( a - b \right)}\left[ \text{ ln }\left| \text{ sec }\left( x - b \right) \right| - \text{ ln } \left| \text{ sec }\left( x - a \right) \right| \right] + C\]
\[ = \frac{1}{\text{ sin }\left( a - b \right)}\left[ \text{ ln }\left| \text{ cos }\left( x - a \right) \right| - \text{ ln }\left| \text{ cos}\left( x - b \right) \right| \right] + C\]
\[ = \frac{1}{\text{ sin }\left( a - b \right)}\left[ \text{ ln }\left| \frac{\text{ cos}\left( x - a \right)}{\text{ cos}\left( x - b \right)} \right| \right] + C\]