Chamemos a integral procurada de \(I\):

\[ I=\int \textrm{e}^{-\theta}\cos(2\theta)\,d\theta. \]

Tomamos \(u=\cos(2\theta)\) e \(dv=\textrm{e}^{-\theta}\,d\theta\). Então

\[ du=-2\operatorname{sen}(2\theta)\,d\theta, \qquad v=-\textrm{e}^{-\theta}. \]

Daí

\[ I = -\textrm{e}^{-\theta}\cos(2\theta) - 2\int \textrm{e}^{-\theta}\operatorname{sen}(2\theta)\,d\theta. \]

Aplicando integração por partes novamente na integral com seno e substituindo de volta, obtemos

\[ I = -\textrm{e}^{-\theta}\cos(2\theta) + 2\textrm{e}^{-\theta}\operatorname{sen}(2\theta) - 4I. \]

Assim, \(5I=\textrm{e}^{-\theta}\bigl(-\cos(2\theta)+2\operatorname{sen}(2\theta)\bigr)\). Portanto,

\[ \colorbox{green!20}{\int \textrm{e}^{-\theta}\cos(2\theta)\,d\theta = \frac{\textrm{e}^{-\theta}}{5} \left(-\cos(2\theta)+2\operatorname{sen}(2\theta)\right)+C}. \]