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Question Number 36728 by a1bgt3@gmail.com last updated on 04/Jun/18

$$theimproperintegral{\int }_0^1\frac{dx}{\sqrt{1-x^2}}convergesto$$

Commented by abdo.msup.com last updated on 05/Jun/18

$$I={lim}_{\xi \to 0}{\int }_{\xi }^1\frac{dx}{\sqrt{1-x^2}}butthechangrment$$$$x=sin\theta give$$$${\int }_{\xi }^1\frac{dx}{\sqrt{1-x^2}}={\int }_{arcsin\left(\xi \right)}^{\frac{\pi }{2}}\frac{cos\theta d\theta }{cos\theta }d\theta$$$$=\frac{\pi }{2}-arcsin\left(\xi \right)\Rightarrow {lim}_{\xi \to 0}{\int }_{\xi }^1\frac{dx}{\sqrt{1-x^2}}=\frac{\pi }{2}$$$$soI=\frac{\pi }{2}.$$

Answered by MJS last updated on 04/Jun/18

$$\underset{0}{\stackrel{1}{\int }}\frac{dx}{\sqrt{1-x^2}}={\left[\arcsin x\right]}_0^1=\frac{\pi }{2}$$$$...\mathrm{nothing}\mathrm{improper}\mathrm{about}\mathrm{this}...$$

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