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