Evaluating the integral $\int_{-\pi/4}^{\pi/4} 11 \sqrt{\sec^2\theta-1}\,d\theta$ Evaluating the integral $\int_{-\pi/4}^{\pi/4} 11 \sqrt{\sec^2\theta-1}\,d\theta$.
Recall that $\tan^2\theta+1=\sec^2\theta \iff \sec^2\theta-1=\tan^2\theta$. So,
\begin{equation*}
\begin{aligned}
\int_{-\pi/4}^{\pi/4} 11 \sqrt{\sec^2\theta-1}\,d\theta
& = 11 \int_{-\pi/4}^{\pi/4} \sqrt{\tan^2\theta}\,d\theta \\
& = 11 \int_{-\pi/4}^{\pi/4} \tan\theta\,d\theta \\
& = 11 \int_{-\pi/4}^{\pi/4} \dfrac{\sin\theta}{\cos\theta}\,d\theta \\
\end{aligned}
\end{equation*}
Let $u=\cos\theta$, $\,du=-\sin\theta\,du$.
Hence
\begin{equation*}
\begin{aligned}
& = -11 \int \dfrac{\,du}{u} \\
& = -11 \left. \ln|\cos \theta|\right]_{-\pi/4}^{\pi/4} \\
& = -11 (\ln|\cos (pi/4)| - \ln|\cos (-pi/4)|) \\
& = -11 (\ln|\dfrac{\sqrt{2}}{2}| - \ln|\dfrac{\sqrt{2}}{2}|) \\
\end{aligned}
\end{equation*}
which equals 0 but the textbook says the answer is $11\ln 2$. Where did I mess up?

\begin{equation*}
\begin{aligned}
\int_{-\pi/4}^{\pi/4} 11 \sqrt{\sec^2\theta-1}\,d\theta
& = 11 \int_{-\pi/4}^{\pi/4} \sqrt{\tan^2\theta}\,d\theta \\
& = 11 \int_{-\pi/4}^{\pi/4} \lvert\tan\theta\rvert\,d\theta \\
& = 2\cdot11\int_{0}^{\pi/4} \tan\theta\,d\theta \\
& = 22[-\log(\cos \theta)]_{0}^{\pi/4}=22(-\log(\cos(\pi/4))+\log(\cos0))  \\
& = 22\frac{\log 2}{2}=11\log 2
\end{aligned}
\end{equation*}
