## Thomas' Calculus 13th Edition

$$\frac{\pi }{{16}}$$
\eqalign{ & \int_0^2 {\frac{{dt}}{{8 + 2{t^2}}}} \cr & = \int_0^2 {\frac{{dt}}{{{{\left( {\sqrt 8 } \right)}^2} + {{\left( {\sqrt 2 t} \right)}^2}}}} \cr & {\text{use the substitution method}}{\text{.}} \cr & u = \sqrt 2 t,{\text{ so that }}du = \sqrt 2 dt \cr & {\text{the new limits on }}t{\text{ are found as follows}} \cr & \,\,\,\,\,\,{\text{If }}t = 2,{\text{ then }}u = \sqrt 2 \left( 2 \right) = 2\sqrt 2 \cr & \,\,\,\,\,\,{\text{If }}t = 0,{\text{ then }}u = \sqrt 2 \left( 0 \right) = 0 \cr & {\text{then}} \cr & \int_0^2 {\frac{{dt}}{{{{\left( {\sqrt 8 } \right)}^2} + {{\left( {\sqrt 2 t} \right)}^2}}}} = \int_0^{2\sqrt 2 } {\frac{{du/\sqrt 2 }}{{{{\left( {\sqrt 8 } \right)}^2} + {u^2}}}} \cr & = \frac{1}{{\sqrt 2 }}\int_0^{2\sqrt 2 } {\frac{{du}}{{{{\left( {2\sqrt 2 } \right)}^2} + {u^2}}}} \cr & {\text{integrate by using the formula }}\int {\frac{{du}}{{{a^2} + {u^2}}} = \frac{1}{a}{{\tan }^{ - 1}}\left( {\frac{u}{a}} \right) + C\,\,\,\left( {{\text{see page 419}}} \right)} \cr & {\text{with }}a = 2\sqrt 2 \cr & = \frac{1}{{\sqrt 2 }}\left( {\frac{1}{{2\sqrt 2 }}ta{n^{ - 1}}\left( {\frac{u}{{\sqrt 8 }}} \right)} \right)_0^{2\sqrt 2 } \cr & = \frac{1}{4}\left( {ta{n^{ - 1}}\left( {\frac{{2\sqrt 2 }}{{2\sqrt 2 }}} \right) - ta{n^{ - 1}}\left( {\frac{0}{{2\sqrt 2 }}} \right)} \right) \cr & = \frac{1}{4}\left( {\frac{\pi }{4} - 0} \right) \cr & = \frac{\pi }{{16}} \cr}