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

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

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

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(B) माना $I = \int_{-\pi / 4}^{\pi / 3}\left|	an \left(x-\frac{\pi}{6}ight)ight| d x$. माना $u = x - \frac{\pi}{6}$,तो $du = dx$. जब $x = -\frac{\pi}{4}$,तब $u = -\frac{5\pi}{12}$. जब $x = \frac{\pi}{3}$,तब $u = \frac{\pi}{6}$. अतः,$I = \int_{-5\pi/12}^{\pi/6} |	an u| du = \int_{-5\pi/12}^{0} -	an u du + \int_{0}^{\pi/6} 	an u du$. $I = [\log |\cos u|]_{-5\pi/12}^{0} + [-\log |\cos u|]_{0}^{\pi/6}$. $I = -\log |\cos(5\pi/12)| - \log |\cos(\pi/6)|$. $\cos(75^\circ) = \frac{\sqrt{6}-\sqrt{2}}{4}$ और $\cos(30^\circ) = \frac{\sqrt{3}}{2}$ का उपयोग करने पर। $I = \log \left( \frac{8}{3\sqrt{2}-\sqrt{6}} ight) = \log(2\sqrt{2}(\sqrt{3}+1))$.

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

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