Single Variable Calculus, Chapter 5, 5.5, Section 5.5, Problem 40

Find the definite integral $\displaystyle \int^{\frac{1}{2}}_{\frac{1}{6}} \csc \pi t \cot \pi t dt$

Let $u = \pi t$, then $du = \pi dt$, so $\displaystyle dt = \frac{du}{\pi}$. When $\displaystyle t = \frac{1}{6}, u = \frac{\pi}{6}$ and when $\displaystyle t = \frac{1}{2}, u = \frac{\pi}{2}$. Thus,


$
\begin{equation}
\begin{aligned}

\int^{\frac{1}{2}}_{\frac{1}{6}} \csc \pi t \cot \pi t dt =& \int^{\frac{1}{2}}_{\frac{1}{6}} \csc u \cot u \frac{du}{\pi}
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\int^{\frac{1}{2}}_{\frac{1}{6}} \csc \pi t \cot \pi t dt =& \frac{1}{\pi} \int^{\frac{1}{2}}_{\frac{1}{6}} \csc u \cot u du
\\
\\
\int^{\frac{1}{2}}_{\frac{1}{6}} \csc \pi t \cot \pi t dt =& \left. \frac{1}{\pi} (- \csc u) \right|^{\frac{1}{2}}_{\frac{1}{6}}
\\
\\
\int^{\frac{1}{2}}_{\frac{1}{6}} \csc \pi t \cot \pi t dt =& \frac{\displaystyle - \csc \frac{\pi}{2}}{\pi} - \frac{\displaystyle \left( - \csc \frac{\pi}{6} \right)}{\pi}
\\
\\
\int^{\frac{1}{2}}_{\frac{1}{6}} \csc \pi t \cot \pi t dt =& \frac{1}{\pi}


\end{aligned}
\end{equation}
$