Answer
$$ - \frac{1}{5}{\tan ^{ - 1}}\left( {\cos 5t} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sin 5t}}{{1 + {{\left( {\cos 5t} \right)}^2}}}dt} \cr
& {\text{Let }}u = \cos 5t,\,\,\,\,du = - 5\sin 5tdt,\,\,\,dt = - \frac{{du}}{{5\sin 5t}} \cr
& {\text{Write the integrand in terms of }}u \cr
& \int {\frac{{\sin 5t}}{{1 + {{\left( {\cos 5t} \right)}^2}}}dt} = \int {\frac{{\sin 5t}}{{1 + {u^2}}}\left( { - \frac{{du}}{{5\sin 5t}}} \right)} \cr
& = - \frac{1}{5}\int {\frac{{du}}{{1 + {u^2}}}} \cr
& {\text{use the formula }}\int {\frac{{du}}{{{a^2} + {u^2}}} = \frac{1}{a}{{\tan }^{ - 1}}\left( {\frac{u}{a}} \right) + C} \cr
& = - \frac{1}{5}{\tan ^{ - 1}}u + C \cr
& \cr
& {\text{Write in terms of }}t;{\text{ substitute }}\cos 5t{\text{ for }}u \cr
& = - \frac{1}{5}{\tan ^{ - 1}}\left( {\cos 5t} \right) + C \cr} $$
