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