Single Variable Calculus, Chapter 5, 5.3, Section 5.3, Problem 32

Find the integral $\displaystyle \int^{\frac{\pi}{4}}_0 \sec \theta \tan \theta d \theta$

Using 2nd Fundamental Theorem of Calculus

$\displaystyle \int^b_a f(x) dx = F(b) - F(a)$, where $F$ is any anti-derivative of $f$.

Let $\displaystyle f(x) = \sec \theta \tan \theta$, then

$\displaystyle F(x) = \sec \theta + C$ or $\displaystyle F(x) = \frac{1}{\cos \theta} + C$


$
\begin{equation}
\begin{aligned}

\int^{\frac{\pi}{4}}_0 \sec \theta \tan \theta =& F\left( \frac{\pi}{4} \right) - F(0)
\\
\\
\int^{\frac{\pi}{4}}_0 \sec \theta \tan \theta =& \frac{1}{\displaystyle \cos \left( \frac{\pi}{4} \right)} + C - \left[ \frac{1}{\cos (0)} + C \right]
\\
\\
\int^{\frac{\pi}{4}}_0 \sec \theta \tan \theta =& \sqrt{2} + C - \frac{1}{1} - C
\\
\\
\int^{\frac{\pi}{4}}_0 \sec \theta \tan \theta =& \sqrt{2} - 1

\end{aligned}
\end{equation}
$