Single Variable Calculus, Chapter 8, 8.2, Section 8.2, Problem 18

Determine the integral $\displaystyle \int \cot^5 \theta \sin^4 \theta d \theta$


$\begin{equation} \begin{aligned} \int \cot^5 \theta \sin^4 \theta d \theta =& \int \frac{\cos^5 \theta}{\sin^5 \theta} \sin^4 \theta d \theta \\ \\ \int \cot^5 \theta \sin^4 \theta d \theta =& \int \frac{\cos^5 \theta}{\sin \theta} d \theta \\ \\ \int \cot^5 \theta \sin^4 \theta d \theta =& \int \frac{\cos^4 \theta}{\sin \theta} d \theta \\ \\ \int \cot^5 \theta \sin^4 \theta d \theta =& \int \frac{(\cos^2 \theta)^2}{\sin \theta} \cos \theta d \theta \qquad \text{Apply Pythagorean Identities } \cos^2 \theta + \sin^2 \theta = 1 \\ \\ \int \cot^5 \theta \sin^4 \theta d \theta =& \int \frac{(1 - \sin^2 \theta)^2}{\sin \theta} \cos \theta d \theta \end{aligned} \end{equation}$


Let $u = \sin \theta$, then $du = \cos \theta d \theta$. Thus,


$\begin{equation} \begin{aligned} \int \frac{(1 - \sin^2 \theta)^2}{\sin \theta} \cos \theta d \theta =& \int \frac{(1 - u^2)^2}{u} du && \\ \\ \int \frac{(1 - \sin^2 \theta)^2}{\sin \theta} \cos \theta d \theta =& \int \left( \frac{1 - 2u^2 + u^4}{u} \right) du && \\ \\ \int \frac{(1 - \sin^2 \theta)^2}{\sin \theta} \cos \theta d \theta =& \int \left( \frac{1}{u} - \frac{2u^2}{u} + \frac{u^4}{u} \right) du && \\ \\ \int \frac{(1 - \sin^2 \theta)^2}{\sin \theta} \cos \theta d \theta =& \int \left( \frac{1}{u} - 2u + u^3 \right) du && \\ \\ \int \frac{(1 - \sin^2 \theta)^2}{\sin \theta} \cos \theta d \theta =& \ln u - \frac{2u^{1 + 1}}{1 + 1} + \frac{u^{3 + 1}}{3 + 1} + c && \\ \\ \int \frac{(1 - \sin^2 \theta)^2}{\sin \theta} \cos \theta d \theta =& \ln U - \frac{\cancel{2}u^2}{\cancel{2}} + \frac{u^4}{4} + c && \\ \\ \int \frac{(1 - \sin^2 \theta)^2}{\sin \theta} \cos \theta d \theta =& \ln u - u^2 + \frac{u^4}{4} + c && \text{Substitute value of } u \\ \\ \int \frac{(1 - \sin^2 \theta)^2}{\sin \theta} \cos \theta d \theta =& \ln (\sin \theta) - (\sin \theta)^2 + \frac{(\sin \theta)^4}{4} + c && \\ \\ \int \frac{(1 - \sin^2 \theta)^2}{\sin \theta} \cos \theta d \theta =& \ln (\sin \theta) - \sin^2 \theta + \frac{\sin^4 \theta}{4} + c \end{aligned} \end{equation}$