Question

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

Answer 1

\[\int\frac{1}{\cos\left( x + a \right) \cos\left( x + b \right)}dx\]
\[\text{Multiplying and Dividing by} \sin\left[ \left( x + b \right) - \left( x + a \right) \right], \text{we get}\]
\[ = \int\frac{1}{\sin\left[ \left( x + b \right) - \left( x + a \right) \right]} \times \frac{\sin\left[ \left( x + b \right) - \left( x + a \right) \right]}{\cos\left( x + a \right) \cos\left( x + b \right)}dx\]
\[ = \int\frac{1}{\sin\left( b - a \right)} \times \frac{\sin\left[ \left( x + b \right) - \left( x + a \right) \right]}{\cos\left( x + a \right) \cos\left( x + b \right)}dx\]

\[ = \frac{1}{\sin\left( b - a \right)}\int\frac{\sin\left( x + b \right)\cos\left( x + a \right) - \sin\left( x + a \right)\cos\left( x + b \right)}{\cos\left( x + a \right) \cos\left( x + b \right)}dx\]
\[ = \frac{1}{\sin\left( b - a \right)}\left[ \int\frac{\sin\left( x + b \right)}{\cos\left( x + b \right)}dx - \int\frac{\sin\left( x + a \right)}{\cos\left( x + a \right)}dx \right]\]
\[ = \frac{1}{\sin\left( b - a \right)}\left[ \int\tan\left( x + b \right)dx - \int\tan\left( x + a \right)dx \right]\]
\[ = \frac{1}{\sin\left( b - a \right)}\left[ \log\left( \sec\left( x + b \right) \right) - \log\left( \sec\left( x + a \right) \right) \right] + c\]
\[ = \frac{1}{\sin\left( b - a \right)}\left[ \log\left( \frac{\sec\left( x + b \right)}{\sec\left( x + a \right)} \right) \right] + c\]

 Hence , \[\int\frac{1}{\cos\left( x + a \right) \cos\left( x + b \right)}dx = \frac{1}{\sin\left( b - a \right)}\left[ \log\left( \frac{\sec\left( x + b \right)}{\sec\left( x + a \right)} \right) \right] + c\]