Calculamos I_{1} e I_{2} separadamente: I_{1}=\displaystyle\int_{0}^{\pi/3}8\sin\varphi\,d\varphi
=8\bigl[-\cos\varphi\bigr]_{0}^{\pi/3}
=8\bigl(-\tfrac12 +1\bigr)
=4. Para I_{2}=\displaystyle\int_{0}^{\pi/3}\operatorname{sec}^{3}\varphi\,\sin\varphi\,d\varphi, fazemos a substituição u=\cos\varphi, du=-\sin\varphi\,d\varphi. Então