Single Variable Calculus, Chapter 5, 5.4, Section 5.4, Problem 34

Find the integrals $\displaystyle \int^{\pi/3}_{\pi/4} \sec \theta \tan \theta d \theta$

$
\begin{equation}
\begin{aligned}
\int \sec \theta \tan \theta d \theta &= \sec \theta + C\\
\\
\int \sec \theta \tan \theta d \theta &= \frac{1}{\cos \theta} + C\\
\\
\int^{\pi/3}_{\pi/4} \sec \theta \tan \theta d \theta &= \frac{1}{\cos \left( \frac{\pi}{3} \right)} + C - \left[ \frac{1}{\cos \left( \frac{\pi}{4} \right)} + C \right]\\
\\
\int^{\pi/3}_{\pi/4} \sec \theta \tan \theta d \theta &= \frac{1}{\frac{1}{2}} + C - \frac{1}{\frac{\sqrt{2}}{2}} - C\\
\\
\int^{\pi/3}_{\pi/4} \sec \theta \tan \theta d \theta &= 2 - \frac{2}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}\\
\\
\int^{\pi/3}_{\pi/4} \sec \theta \tan \theta d \theta &= 2 - \frac{\cancel{2}\sqrt{2}}{\cancel{2}}\\
\\
\int^{\pi/3}_{\pi/4} \sec \theta \tan \theta d \theta &= 2 - \sqrt{2}
\end{aligned}
\end{equation}
$