Single Variable Calculus, Chapter 8, 8.2, Section 8.2, Problem 40

Determine the integral $\displaystyle \int \csc^4 x \cot^6 x dx$


$\begin{equation} \begin{aligned} \int \csc^4 x \cot^6 x dx =& \int \csc^2 x \csc^2 x \cot^6 x dx \qquad \text{Apply Trigonometric Identities } \csc^2 x = 1 + \cot^2 x \\ \\ \int \csc^4 x \cot^6 x dx =& \int \csc^2 x (1 + \cot^2 x) \cot^6 x dx \\ \\ \int \csc^4 x \cot^6 x dx =& \int \csc^2 x (\cot^6 x + \cot^8 x) dx \end{aligned} \end{equation}$


Let $u = \cot x$, then $du = - \csc^2 x dx$, so $\csc^2 x dx = -du$. Therefore,


$\begin{equation} \begin{aligned} \int \csc^2 x (\cot^6 x + \cot^8 x) dx =& \int (u^6 + u^8) \cdot -du \\ \\ \int \csc^2 x (\cot^6 x + \cot^8 x) dx =& - \int (u^6 + u^8) du \\ \\ \int \csc^2 x (\cot^6 x + \cot^8 x) dx =& - \left( \frac{u^{6 + 1}}{6 + 1} + \frac{u^{8 + 1}}{8 + 1} \right) + c \\ \\ \int \csc^2 x (\cot^6 x + \cot^8 x) dx =& \frac{-u^7}{7} - \frac{u^9}{9} + c \\ \\ \int \csc^2 x (\cot^6 x + \cot^8 x) dx =& \frac{- (\cot x)^7}{7} - \frac{(\cot x)^9}{9} + c \\ \\ \int \csc^2 x (\cot^6 x + \cot^8 x) dx =& - \frac{\cot^7x}{7} - \frac{\cot^9 x}{9} + c \end{aligned} \end{equation}$