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Question Number 124644 by benjo_mathlover last updated on 05/Dec/20

$$\underset{2/\sqrt{3}}{\stackrel{2}{\int }}\frac{\cos \left({\sec }^{-1}x\right)}{x\sqrt{x^2-1}}dx$$$$\underset{\sqrt{2}}{\stackrel{2}{\int }}\frac{\sec {}^2\left({\sec }^{-1}x\right)}{x\sqrt{x^2-1}}dx$$

Answered by liberty last updated on 05/Dec/20

$$\left(1\right)\int \underset{\frac{2}{\sqrt{3}}}{\stackrel{2}{}}\frac{\cos \left({\sec }^{-1}x\right)}{x\sqrt{x^2-1}}dx$$$$put{\sec }^{-1}x=j,x=\sec janddx=\sec j\tan jdj$$$$whereupperlimitj=\frac{\pi }{3}andlowerlimitj=\frac{\pi }{6}$$$$\int {\stackrel{\pi /3}{}}_{\pi /6}\frac{\cos j}{\sec j\tan j}\left(\sec j\tan j\right)dj$$$$I={\left[\sin j\right]}_{\frac{\pi }{6}}^{\frac{\pi }{3}}=\frac{\sqrt{3}}{2}-\frac{1}{2}=\frac{1}{2}(\sqrt{3}-1)$$

Answered by Dwaipayan Shikari last updated on 05/Dec/20

$${\int }_{\sqrt{2}}^2\frac{{sec}^2\left({sec}^{-1}x\right)}{x\sqrt{x^2-1}}dx{sec}^{-1}x=t\Rightarrow \frac{1}{x\sqrt{x^2-1}}=\frac{dt}{dx}$$$$={\int }_{\frac{\pi }{4}}^{\frac{\pi }{3}}{sec}^2\left(t\right)dt={\left[tan\left(t\right)\right]}_{\frac{\pi }{4}}^{\frac{\pi }{3}}=\sqrt{3}-1$$

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