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Question Number 145588 by naka3546 last updated on 06/Jul/21

Answered by chengulapetrom last updated on 06/Jul/21

$$let2sec\theta =x$$$$2sec\theta tan\theta d\theta =dx$$$$=2\int \frac{2{tan}^2\theta sec\theta }{2sec\theta }d\theta$$$$=2\int {tan}^2\theta d\theta$$$$=\int {sec}^2\theta d\theta -\int d\theta$$$$=2tan\theta -2\theta +C$$$$=2tan\left({sec}^{-1}\left(\frac{x}{2}\right)\right)-2{sec}^{-1}\left(\frac{x}{2}\right)+C$$

Answered by qaz last updated on 06/Jul/21

$$\int \frac{\sqrt{{\mathrm{x}}^2-4}}{\mathrm{x}}\mathrm{dx}$$$$=2\int \frac{\tan \mathrm{u}}{2\sec \mathrm{u}}\cdot 2\sec \mathrm{u}\cdot \tan \mathrm{udu}..........\mathrm{x}=2\sec \mathrm{u}$$$$=2\int \tan {}^2\mathrm{udu}$$$$=2\int (\sec {}^2\mathrm{u}-1)\mathrm{du}$$$$=2\tan \mathrm{u}-2\mathrm{u}+\mathrm{C}$$$$=\sqrt{{\mathrm{x}}^2-4}-2\sec {}^{-1}\frac{\mathrm{x}}{2}+\mathrm{C}$$

Answered by puissant last updated on 06/Jul/21

$$\mathrm{x}=2\sec \theta \Rightarrow \theta ={\sec }^{-1}\left(\frac{\mathrm{x}}{2}\right)$$$$\Rightarrow \mathrm{dx}=2\sec \theta \tan \theta \mathrm{d}\theta$$$$\Rightarrow \mathrm{I}=\int \frac{\sqrt{\left({2\sec }^2\theta \right)-4}}{2\sec \theta }\left(2\sec \theta \tan \theta \right)\mathrm{d}\theta$$$$=2\int \sqrt{{\sec }^2\theta -1}\tan \theta \mathrm{d}\theta$$$$=2\int {\tan }^2\theta \mathrm{d}\theta =2\int {\tan }^2\theta +1-1\mathrm{d}\theta$$$$\Rightarrow \mathrm{I}=2\tan \theta -2\theta +\mathrm{k}$$$$\mathrm{I}=2\tan \left({\sec }^{-1}\left(\frac{\mathrm{x}}{2}\right)\right)-{2\sec }^{-1}\left(\frac{\mathrm{x}}{2}\right)+\mathrm{k}..$$

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