Integrals — Class 12 Maths Solution

Let $I = \int {\cfrac{1}{{\cos \left( {x + a} \right)\cos \left( {x + b} \right)}}dx}$

$= \cfrac{1}{{\sin \left( {a - b} \right)}}\int {\cfrac{{\sin \left[ {\left( {x + a} \right) - \left( {x + b} \right)} \right]}}{{\cos \left( {x + a} \right)\cos \left( {x + b} \right)}}dx}$

$= \cfrac{1}{{\sin \left( {a - b} \right)}}\int {\cfrac{{\sin \left( {x + a} \right)\cos \left( {x + b} \right) - \cos \left( {x + a} \right)\sin \left( {x + b} \right)}}{{\cos \left( {x + a} \right)\cos \left( {x + b} \right)}}dx}$

$= \cfrac{1}{{\sin \left( {a - b} \right)}}\left[ {\log \left| {\sec \left( {x + a} \right)} \right| - \log \left| {\sec \left( {x + b} \right)} \right|} \right] + C$

$= \cfrac{1}{{\sin \left( {a - b} \right)}}\log \left| {\cfrac{{\sec \left( {x + a} \right)}}{{\sec \left( {x + b} \right)}}} \right| + C$

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