There really isn’t too much to this problem. Just recall that the formula from the notes assumes that \(x = f\left( t \right)\) and \(y = g\left( t \right)\). So, the area under the curve is,

\[\begin{align*}A & = \int_{0}^{\pi }{{\left( {4 + \sin \left( t \right)} \right)\left( {3{{\cos }^2}\left( t \right)\sin \left( t \right)} \right)\,dt}}\\ & = \int_{0}^{\pi }{{12{{\cos }^2}\left( t \right)\sin \left( t \right) + 3{{\cos }^2}\left( t \right){{\sin }^2}\left( t \right)\,dt}}\\ & = \int_{0}^{\pi }{{12{{\cos }^2}\left( t \right)\sin \left( t \right) + 3{{\left[ {\frac{1}{2}\sin \left( {2t} \right)} \right]}^2}\,dt}}\\ & = \int_{0}^{\pi }{{12{{\cos }^2}\left( t \right)\sin \left( t \right) + \frac{3}{4}{{\sin }^2}\left( {2t} \right)\,dt}}\\ & = \int_{0}^{\pi }{{12{{\cos }^2}\left( t \right)\sin \left( t \right) + \frac{3}{8}\left( {1 - cos\left( {4t} \right)} \right)\,dt}}\\ & = \left. {\left( { - 4{{\cos }^3}\left( t \right) + \frac{3}{8}t - \frac{3}{{32}}\sin \left( {4t} \right)} \right)} \right|_{0}^{\pi } = \require{bbox} \bbox[2pt,border:1px solid black]{{8 + \frac{3}{8}\pi }}\end{align*}\]

You did recall how to do all the trig manipulations and trig integrals to do this integral correct? If not you should go back to the Integrals Involving Trig Functions section to do some review/problems.