int_{-pi / 4}^{pi / 3}left|tan left(x-frac{pi | vedclass

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(B) Let $I = \int_{-\pi / 4}^{\pi / 3}\left|	an \left(x-\frac{\pi}{6}ight)ight| d x$. Let $u = x - \frac{\pi}{6}$,then $du = dx$. When $x = -\fra

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$\log (2 \sqrt{2}(\sqrt{3}+1))$

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(B) Let $I = \int_{-\pi / 4}^{\pi / 3}\left|	an \left(x-\frac{\pi}{6}ight)ight| d x$. Let $u = x - \frac{\pi}{6}$,then $du = dx$. When $x = -\frac{\pi}{4}$,$u = -\frac{\pi}{4} - \frac{\pi}{6} = -\frac{5\pi}{12}$. When $x = \frac{\pi}{3}$,$u = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$. So,$I = \int_{-5\pi/12}^{\pi/6} |	an u| du = \int_{-5\pi/12}^{0} |	an u| du + \int_{0}^{\pi/6} |	an u| du$. Since $	an u  0$ for $u \in (0, \frac{\pi}{6}]$,we have: $I = \int_{-5\pi/12}^{0} -	an u du + \int_{0}^{\pi/6} 	an u du$. $I = [\log |\cos u|]_{-5\pi/12}^{0} + [-\log |\cos u|]_{0}^{\pi/6}$. $I = (\log 1 - \log |\cos(-5\pi/12)|) - (\log |\cos(\pi/6)| - \log 1)$. $I = -\log |\cos(5\pi/12)| - \log |\cos(\pi/6)| = -\log |\cos(75^\circ)| - \log |\cos(30^\circ)|$. Using $\cos(75^\circ) = \cos(45^\circ+30^\circ) = \frac{\sqrt{6}-\sqrt{2}}{4}$ and $\cos(30^\circ) = \frac{\sqrt{3}}{2}$. $I = -\log \left( \frac{\sqrt{6}-\sqrt{2}}{4} \cdot \frac{\sqrt{3}}{2} ight) = -\log \left( \frac{3\sqrt{2}-\sqrt{6}}{8} ight) = \log \left( \frac{8}{3\sqrt{2}-\sqrt{6}} ight)$. Rationalizing the denominator: $\frac{8}{\sqrt{6}(\sqrt{3}-1)} \cdot \frac{\sqrt{3}+1}{\sqrt{3}+1} = \frac{8(\sqrt{3}+1)}{\sqrt{6}(3-1)} = \frac{8(\sqrt{3}+1)}{2\sqrt{6}} = \frac{4(\sqrt{3}+1)}{\sqrt{6}} = 2\sqrt{2}(\sqrt{3}+1)$. Thus,$I = \log(2\sqrt{2}(\sqrt{3}+1))$.

$\int_{-\pi / 4}^{\pi / 3}\left|\tan \left(x-\frac{\pi}{6}\right)\right| d x=$

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