## Thomas' Calculus 13th Edition

$$\frac{\pi }{{12}}$$
\eqalign{ & \int_{\pi /6}^{\pi /4} {\frac{{{{\csc }^2}xdx}}{{1 + {{\left( {\cot x} \right)}^2}}}} \cr & {\text{use the substitution method}}{\text{.}} \cr & u = \cot x,{\text{ so that }}du = - {\csc ^2}xdx \cr & {\text{the new limits on }}t{\text{ are found as follows}} \cr & \,\,\,\,\,\,{\text{If }}x = \pi /4,{\text{ then }}u = \cot \left( {\pi /4} \right) = 1 \cr & \,\,\,\,\,\,{\text{If }}x = \pi /6,{\text{ then }}u = \cot \left( {\pi /6} \right) = \sqrt 3 \cr & {\text{then}} \cr & \int_{\pi /6}^{\pi /4} {\frac{{{{\csc }^2}xdx}}{{1 + {{\left( {\cot x} \right)}^2}}}} = \int_{\sqrt 3 }^1 {\frac{{ - du}}{{1 + {u^2}}}} \cr & {\text{use integral properties}} \cr & = \int_1^{\sqrt 3 } {\frac{{du}}{{1 + {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 = 1 \cr & = \left( {{{\tan }^{ - 1}}\left( u \right)} \right)_1^{\sqrt 3 } + C \cr & = {\tan ^{ - 1}}\left( {\sqrt 3 } \right) - {\tan ^{ - 1}}\left( 1 \right) \cr & = \frac{\pi }{3} - \frac{\pi }{4} \cr & = \frac{\pi }{{12}} \cr}