## Thomas' Calculus 13th Edition

$$2\sqrt 3 - \ln \left( {2 - \sqrt 3 } \right)$$
\eqalign{ & \int_{ - \pi /3}^0 {2{{\sec }^3}x} dx \cr & = 2\int_{ - \pi /3}^0 {{{\sec }^3}x} dx \cr & \cr & {\text{Integrate by parts using}}{\text{,}} \cr & u = \sec x,\,\,\,\,du = \sec x\tan x \cr & dv = {\sec ^2}x,\,\,\,\,\,\,v = \tan x \cr & {\text{using integration by parts formula}} \cr & \int {{{\sec }^3}x} dx = \sec x\tan x - \int {\left( {\tan x} \right)\left( {\sec x\tan x} \right)} dx \cr & \int {{{\sec }^3}x} dx = \sec x\tan x - \int {\left( {{{\tan }^2}x} \right)\sec x} dx \cr & \int {{{\sec }^3}x} dx = \sec x\tan x - \int {\left( {{{\sec }^2}x - 1} \right)\sec x} dx \cr & \int {{{\sec }^3}x} dx = \sec x\tan x - \int {{{\sec }^3}x} dx + \int {\sec x} dx \cr & 2\int {{{\sec }^3}x} dx = \sec x\tan x + \int {\sec x} dx \cr & \int {{{\sec }^3}x} dx = \frac{1}{2}\sec x\tan x + \frac{1}{2}\int {\sec x} dx \cr & {\text{where }}\int {\sec x} dx = \ln \left| {\sec x + \tan x} \right| + C \cr & \int {{{\sec }^3}x} dx = \frac{1}{2}\sec x\tan x + \frac{1}{2}\ln \left| {\sec x + \tan x} \right| + C \cr & {\text{Then}} \cr & 2\int_{ - \pi /3}^0 {{{\sec }^3}x} dx = 2\left( {\frac{1}{2}\sec x\tan x + \frac{1}{2}\ln \left| {\sec x + \tan x} \right|} \right)_{ - \pi /3}^0 \cr & = \left( {\sec x\tan x + \ln \left| {\sec x + \tan x} \right|} \right)_{ - \pi /3}^0 \cr & = \left( {\sec \left( 0 \right)\tan \left( 0 \right) + \ln \left| {\sec 0 + \tan 0} \right|} \right) \cr & - \left( {\sec \left( { - \frac{\pi }{3}} \right)\tan \left( { - \frac{\pi }{3}} \right) + \ln \left| {\sec \left( { - \frac{\pi }{3}} \right) + \tan \left( { - \frac{\pi }{3}} \right)} \right|} \right) \cr & {\text{simplifying}} \cr & = \left( {0 + \ln \left| 1 \right|} \right) - \left( {2\left( { - \sqrt 3 } \right) + \ln \left| {2 - \sqrt 3 } \right|} \right) \cr & = 2\sqrt 3 - \ln \left( {2 - \sqrt 3 } \right) \cr}