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Question Number 106964 by Ar Brandon last updated on 08/Aug/20

$${\int }_0^{\pi }\frac{{\sec }^2\mathrm{x}}{\sqrt{1-{\tan }^2\mathrm{x}}}\mathrm{dx}$$

Commented by Her_Majesty last updated on 08/Aug/20

$$Ibelievethisintegralisdivergent$$$$mustsleepnowtokeepmymajesticbeauty$$$$willtrytoshowlater$$

Commented by Dwaipayan Shikari last updated on 08/Aug/20

$$\mathrm{probably}$$

Commented by Dwaipayan Shikari last updated on 08/Aug/20

Commented by Ar Brandon last updated on 08/Aug/20

��Alright her majesty ��

Answered by bemath last updated on 08/Aug/20

$${}^{@bemath@}$$$$letu=\tan x\to \{\begin{array}{l}du=\sec {}^{2}xdx\\ \to \{\begin{array}{l}x=0\to u=0\\ x=\pi \to u=0\end{array}\end{array}$$$$I=\underset{0}{\stackrel{0}{\int }}\frac{du}{\sqrt{1-u^2}}=0$$

Answered by 1549442205PVT last updated on 08/Aug/20

$$\mathrm{I}={\int }_0^{\pi }\frac{\mathrm{d}\left(\mathrm{tanx}\right)}{\sqrt{1-{\tan }^2\mathrm{x}}}\underset{\mathrm{tanx}=\mathrm{u}}{=}{\int }_0^1\frac{\mathrm{du}}{\sqrt{1-{\mathrm{u}}^2}}={\sin }^{-1}\mathrm{u}{\mid }_0^0$$$$=0-0=0$$

Answered by Dwaipayan Shikari last updated on 08/Aug/20

$${\int }_0^{\pi }\frac{dt}{\sqrt{1-t^2}}tanx=t,{sec}^{2}x=\frac{dt}{dx}$$$${\left[{sin}^{-1}\left(tanx\right)\right]}_0^{\pi }=0$$

Commented by Ar Brandon last updated on 08/Aug/20

$$\mathrm{Thank}{\mathrm{y}}^{\prime }\mathrm{all}.\mathrm{I}\mathrm{really}\mathrm{needed}\mathrm{to}\mathrm{verify}\mathrm{something}.$$$$\mathrm{And}{\mathrm{you}}^{\prime }\mathrm{ve}\mathrm{been}\mathrm{of}\mathrm{great}\mathrm{help}.$$

Answered by Ar Brandon last updated on 08/Aug/20

$$\mathcal{I}={\int }_0^{\frac{\pi }{2}}\frac{{\sec }^2\mathrm{x}}{\sqrt{1-{\tan }^2\mathrm{x}}}\mathrm{dx}+{\int }_{\frac{\pi }{2}}^{\pi }\frac{{\sec }^2\mathrm{x}}{\sqrt{1-{\tan }^2\mathrm{x}}}\mathrm{dx}$$$$={\int }_0^{\frac{\pi }{2}}\frac{{\sec }^2\mathrm{x}}{\sqrt{1-{\tan }^2\mathrm{x}}}\mathrm{dx}-{\int }_0^{\frac{\pi }{2}}\frac{{\sec }^2\mathrm{x}}{\sqrt{1-{\tan }^2\mathrm{x}}}\mathrm{dx}=0$$

Answered by Her_Majesty last updated on 08/Aug/20

$$y=\frac{{sec}^{2}x}{\sqrt{1-{tan}^{2}x}}=\frac{1}{\mid cosx\mid \sqrt{cos2x}}$$$$for0⩽x⩽\pi yisdefinedfor0⩽x<\frac{\pi }{4}and$$$$\frac{3\pi }{4}<y⩽\pi \Rightarrow theintegralonlyexistsifthe$$$$imaginary/complexpartscancelout$$$$but{let}^{\prime }stry$$$$\underset{0}{\stackrel{\pi }{\int }}\frac{dx}{\mid cosx\mid \sqrt{cos2x}}=2\underset{0}{\stackrel{\pi /2}{\int }}\frac{dx}{cosx\sqrt{cos2x}}$$$$witht=tanxweget$$$$\underset{0}{\stackrel{\pi /2}{\int }}\frac{dx}{cosx\sqrt{cos2x}}=\underset{0}{\stackrel{+\mathrm{\infty }}{\int }}\frac{dt}{\sqrt{1-t^2}}={\left[arcsin\left(t\right)\right]}_0^{+\mathrm{\infty }}=$$$$=\underset{t\to +\mathrm{\infty }}{limarcsin}\left(t\right)whichis\frac{\pi }{2}-i\mathrm{\infty }(trysome$$$$valuesliket={10}^{10},{10}^{20}...$$$$\Rightarrow theintegralisdivergent$$

Commented by Ar Brandon last updated on 08/Aug/20

OK Sir, thanks

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