Single Variable Calculus, Chapter 8, 8.2, Section 8.2, Problem 38

Determine the integral $\displaystyle \int^{\frac{\pi}{2}}_{\frac{\pi}{4}} \cot^3 x dx$


$\begin{equation} \begin{aligned} \int^{\frac{\pi}{2}}_{\frac{\pi}{4}} \cot^3 x dx =& \int^{\frac{\pi}{2}}_{\frac{\pi}{4}} \cot x \cot^2 x dx \qquad \text{Apply Trigonometric Identities } \csc^2 x = 1 + \cot^2 x \\ \\ \int^{\frac{\pi}{2}}_{\frac{\pi}{4}} \cot^3 x dx =& \int^{\frac{\pi}{2}}_{\frac{\pi}{4}} \cot x (\csc^2 x - 1) dx \\ \\ \int^{\frac{\pi}{2}}_{\frac{\pi}{4}} \cot^3 x dx =& \int^{\frac{\pi}{2}}_{\frac{\pi}{4}} (\cot x \csc^2 x - \cot x) dx \\ \\ \int^{\frac{\pi}{2}}_{\frac{\pi}{4}} \cot^3 x dx =& \int^{\frac{\pi}{2}}_{\frac{\pi}{4}} \cot x \csc^2 x dx - \int^{\frac{\pi}{2}}_{\frac{\pi}{4}}\cot x dx \end{aligned} \end{equation}$


We integrate the equation term by term

@ 1st term

$\displaystyle \int^{\frac{\pi}{2}}_{\frac{\pi}{4}} \cot x \csc^2 x dx$

Let $u = \cot x$, then $du = - \csc^2 x dx$, so $\csc^2 x dx = -du$. When $\displaystyle x = \frac{\pi}{4}, u = 1$ and when $\displaystyle x = \frac{\pi}{2}, u = 0$. Therefore,


$\begin{equation} \begin{aligned} \int^{\frac{\pi}{2}}_{\frac{\pi}{4}} \cot x \csc^2 x dx =& \int^0_1 u \cdot -du \\ \\ \int^{\frac{\pi}{2}}_{\frac{\pi}{4}} \cot x \csc^2 x dx =& - \int^0_1 u du \\ \\ \int^{\frac{\pi}{2}}_{\frac{\pi}{4}} \cot x \csc^2 x dx =& - \left[ \frac{u^{1 + 1}}{1 + 1} \right]^0_1 \\ \\ \int^{\frac{\pi}{2}}_{\frac{\pi}{4}} \cot x \csc^2 x dx =& - \left[ \frac{u^2}{2} \right]^0_1 \\ \\ \int^{\frac{\pi}{2}}_{\frac{\pi}{4}} \cot x \csc^2 x dx =& \frac{-(0)^2}{2} + \frac{(1)^2}{2} \\ \\ \int^{\frac{\pi}{2}}_{\frac{\pi}{4}} \cot x \csc^2 x dx =& \frac{1}{2} \end{aligned} \end{equation}$


@ 2nd term


$\begin{equation} \begin{aligned} \int^{\frac{\pi}{2}}_{\frac{\pi}{4}} \cot x dx =& \left[ \ln (\sin x) \right]^{\frac{\pi}{2}}_{\frac{\pi}{4}} \\ \\ \int^{\frac{\pi}{2}}_{\frac{\pi}{4}} \cot x dx =& \ln \left( \sin \frac{\pi}{2} \right) - \ln \left( \sin \frac{\pi}{4} \right) \\ \\ \int^{\frac{\pi}{2}}_{\frac{\pi}{4}} \cot x dx =& \ln (1) = \ln \left( \frac{\sqrt{2}}{2} \right) \\ \\ \int^{\frac{\pi}{2}}_{\frac{\pi}{4}} \cot x dx =& - \ln \left( \frac{\sqrt{2}}{2} \right) \end{aligned} \end{equation}$


Combine the results of the integration term by term


$\begin{equation} \begin{aligned} \int^{\frac{\pi}{2}}_{\frac{\pi}{4}} \cot x^3 dx =& \frac{1}{2} - \left( - \ln \left( \frac{\sqrt{2}}{2} \right) \right) \\\\ \int^{\frac{\pi}{2}}_{\frac{\pi}{4}} \cot x^3 dx =& \frac{1}{2} + \ln \left( \frac{\sqrt{2}}{2} \right) \end{aligned} \end{equation}$