## Calculus: Early Transcendentals (2nd Edition)

$$\frac{\pi }{{36}} + \frac{1}{{24}} - \frac{{\sqrt 3 + \pi }}{{48}}$$
\eqalign{ & \int_6^{6\sqrt 3 } {\frac{{{z^2}}}{{{{\left( {{z^2} + 36} \right)}^2}}}dz} \cr & z = 6\tan \theta ,{\text{ }}dz = 6{\sec ^2}\theta d\theta \cr & {\text{new limits of integration}} \cr & \theta = {\tan ^{ - 1}}\left( {z/6} \right) \cr & x = 6\sqrt 3 ,{\text{ }}\theta = \pi /3 \cr & x = 6,{\text{ }}\theta = \pi /4 \cr & {\text{we have}} \cr & \int_6^{6\sqrt 3 } {\frac{{{z^2}}}{{{{\left( {{z^2} + 36} \right)}^2}}}dz} = \int_{\pi /4}^{\pi /3} {\frac{{36{{\tan }^2}\theta }}{{{{\left( {36{{\tan }^2}\theta + 36} \right)}^2}}}\left( {6{{\sec }^2}\theta d\theta } \right)} \cr & = \int_{\pi /4}^{\pi /3} {\frac{{36{{\tan }^2}\theta }}{{{{36}^2}{{\left( {{{\tan }^2}\theta + 1} \right)}^2}}}\left( {6{{\sec }^2}\theta d\theta } \right)} \cr & {\text{pythagorean identity}} \cr & = \int_{\pi /4}^{\pi /3} {\frac{{{{\tan }^2}\theta }}{{6{{\left( {{{\sec }^2}\theta } \right)}^2}}}\left( {{{\sec }^2}\theta d\theta } \right)} = \frac{1}{6}\int_{\pi /4}^{\pi /3} {{{\sin }^2}\theta d\theta } \cr & = \frac{1}{6}\int_{\pi /4}^{\pi /3} {\frac{{1 - \cos 2\theta }}{2}d\theta } \cr & {\text{integrating}} \cr & = \frac{1}{6}\left[ {\frac{\theta }{2} - \frac{1}{4}\sin 2\theta } \right]_{\pi /4}^{\pi /3} \cr & {\text{evaluate limits}} \cr & = \frac{1}{6}\left[ {\frac{\pi }{6} - \frac{1}{4}\sin \frac{{2\pi }}{3}} \right] - \frac{1}{6}\left[ {\frac{\pi }{8} - \frac{1}{4}\sin \frac{\pi }{2}} \right] \cr & {\text{simplify}} \cr & = \frac{1}{6}\left[ {\frac{\pi }{6} - \frac{1}{4}\left( {\frac{{\sqrt 3 }}{2}} \right)} \right] - \frac{1}{6}\left[ {\frac{\pi }{8} - \frac{1}{4}\left( 1 \right)} \right] \cr & = \frac{\pi }{{36}} - \frac{{\sqrt 3 }}{{48}} - \frac{\pi }{{48}} + \frac{1}{{24}} \cr & = \frac{\pi }{{36}} + \frac{1}{{24}} - \frac{{\sqrt 3 + \pi }}{{48}} \cr}