Chapter 7 - Integration Techniques - 7.4 Trigonometric Substitutions - 7.4 Exercises - Page 539: 75

$$A = 25\left( {\sqrt 3 - \ln \sqrt {2 + \sqrt 3 } } \right)$$

$$\eqalign{ & {\text{Let }}f\left( x \right) = {\left( {{x^2} - 25} \right)^{1/2}},\,\,\,\,\left[ {5,10} \right] \cr & {\text{The area under de curve is given by}} \cr & A = \int_5^{10} {{{\left( {{x^2} - 25} \right)}^{1/2}}} dx \cr & {\text{The integrand contains the form }}{x^2} - {a^2} \cr & {x^2} - 25 \to a = 5 \cr & {\text{Use the change of variable }}x = a\sec \theta \cr & x = 5\sec \theta ,\,\,\,dx = 5\sec \theta \tan \theta d\theta \cr & = \int {{{\left( {25{{\sec }^2}\theta - 25} \right)}^{1/2}}\left( {5\sec \theta \tan \theta } \right)d\theta } \cr & = \int {5{{\left( {{{\sec }^2}\theta - 1} \right)}^{1/2}}\left( {5\sec \theta \tan \theta } \right)d\theta } \cr & = 25\int {{{\left( {{{\tan }^2}\theta } \right)}^{1/2}}\left( {\sec \theta \tan \theta } \right)d\theta } \cr & = 25\int {{{\tan }^2}\theta \sec \theta } d\theta \cr & = 25\int {\left( {{{\sec }^2}\theta - 1} \right)\sec \theta } d\theta \cr & = 25\int {\left( {{{\sec }^3}\theta - \sec \theta } \right)} d\theta \cr & \cr & {\text{Integrate using the reduction formula}} \cr & = \frac{{25}}{2}\left( {\sec \theta \tan \theta + \ln \left| {\sec \theta + \tan \theta } \right|} \right) - \frac{{25}}{2}\ln \left| {\sec \theta + \tan \theta } \right| + C \cr & = \frac{{25}}{2}\sec \theta \tan \theta - \frac{{25}}{2}\ln \left| {\sec \theta + \tan \theta } \right| + C \cr & {\text{Write in terms of }}x \cr & = \frac{{25}}{2}\left( {\frac{x}{5}} \right)\left( {\frac{{\sqrt {{x^2} - 25} }}{5}} \right) - \frac{{25}}{2}\ln \left| {\frac{x}{5} + \frac{{\sqrt {{x^2} - 25} }}{5}} \right| + C \cr & = \frac{{x\sqrt {{x^2} - 25} }}{2} - \frac{{25}}{2}\ln \left| {\frac{{x + \sqrt {{x^2} - 25} }}{5}} \right| + C \cr & \cr & ,{\text{then}} \cr & A = \int_5^{10} {{{\left( {{x^2} - 25} \right)}^{1/2}}} dx \cr & A = \frac{{10\sqrt {{{10}^2} - 25} }}{2} - \frac{{25}}{2}\ln \left| {\frac{{10 + \sqrt {{{10}^2} - 25} }}{5}} \right| - \frac{0}{2} - \frac{{25}}{2}\ln \left| {\frac{{5 + \sqrt 0 }}{5}} \right| \cr & A = 5\sqrt {75} - \frac{{25}}{2}\ln \left| {2 + \sqrt 3 } \right| \cr & A = 25\sqrt 3 - \frac{{25}}{2}\ln \left( {2 + \sqrt 3 } \right) \cr & A = 25\left( {\sqrt 3 - \frac{1}{2}\ln \left( {2 + \sqrt 3 } \right)} \right) \cr & A = 25\left( {\sqrt 3 - \ln \sqrt {2 + \sqrt 3 } } \right) \cr} $$