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Question Number 143087 by bramlexs22 last updated on 10/Jun/21

Answered by Ar Brandon last updated on 10/Jun/21

$$\mathrm{x}=\sec \vartheta$$$$\mathrm{I}={\int }_{\frac{2\pi }{3}}^{\frac{5\pi }{6}}\frac{\sec \vartheta \tan \vartheta }{\sec \vartheta \sqrt{{\sec }^2\vartheta -1}}\mathrm{d}\vartheta ={\int }_{\frac{2\pi }{3}}^{\frac{5\pi }{6}}\frac{\tan \vartheta }{\sqrt{{\tan }^2\vartheta }}\mathrm{d}\vartheta$$$$={\left[\frac{\tan \vartheta }{\mid \tan \vartheta \mid }\vartheta \right]}_{2\pi /3}^{5\pi /6}=-\frac{5\pi }{6}-(-\frac{2\pi }{3})=-\frac{\pi }{6}$$

Answered by EDWIN88 last updated on 10/Jun/21

$$\iff \mathrm{let}\mathrm{x}=\sec \mathrm{h}\to \{\begin{array}{l}\mathrm{x}=-\frac{2}{\sqrt{3}};\mathrm{h}=\frac{5\pi }{6}\\ \mathrm{x}=-2;\mathrm{h}=\frac{2\pi }{3}\end{array}$$$$\mathrm{I}={\int }_{2\pi /3}^{5\pi /6}\frac{\cos \mathrm{h}}{-\tan \mathrm{h}}\sec \mathrm{h}\tan \mathrm{h}\mathrm{dh}$$$$\mathrm{I}=-{\left[\mathrm{h}\right]}_{2\pi /3}^{5\pi /6}=-\frac{\pi }{6}◻$$

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