[tex] \large \mathcal{ANSWER:} [/tex]
[tex] \boxed{\dfrac{\sin^4 y}{4} - \dfrac{\sin^6 y}{6} + C} [/tex]
[tex] \large \mathcal{SOLUTION:} [/tex]
[tex] \begin{array}{l} \displaystyle \int \cos^3 y \sin^3 y\:dy \\ = \displaystyle \int (\cos^2 y)\cos y\sin^3 y\:dy \\ = \displaystyle \int (1 - \sin^2y)\sin^3 y \cos y\:dy \\ = \displaystyle \int \underbrace{(\sin^3 y - \sin^5 y)}_{\footnotesize u^3 - u^5}\negthickspace \underbrace{\cos y\: dy}_{\begin{aligned} \scriptsize\text{Let}\: u &\scriptsize = \sin y \\ \scriptsize du &\scriptsize= \cos y\:dy \end{aligned}} \\ = \dfrac{\sin^4 y}{4} - \dfrac{\sin^6 y}{6} + C \\ \\ \\ \footnotesize \text{Use the trigonometric identity} \sin^2\theta + \cos^2\theta = 1.\end{array} [/tex]
[tex] \texttt \color{cyan} {\#CarryOnLearning} [/tex]
