Question Number 36728 by a1bgt3@gmail.com last updated on 04/Jun/18
$${the}\:{improper}\:{integral}\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:{converges}\:{to} \\ $$
Commented by abdo.msup.com last updated on 05/Jun/18
$${I}\:={lim}_{\xi\rightarrow\mathrm{0}} \:\int_{\xi} ^{\mathrm{1}} \:\:\:\frac{{dx}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\:{but}\:{the}\:{changrment} \\ $$$${x}\:={sin}\theta\:\:{give}\: \\ $$$$\:\int_{\xi} ^{\mathrm{1}} \:\:\:\frac{{dx}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:=\:\int_{{arcsin}\left(\xi\right)} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{cos}\theta\:{d}\theta}{{cos}\theta}{d}\theta \\ $$$$=\frac{\pi}{\mathrm{2}}\:−{arcsin}\left(\xi\right)\:\Rightarrow{lim}_{\xi\rightarrow\mathrm{0}} \int_{\xi} ^{\mathrm{1}} \:\:\frac{{dx}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:=\frac{\pi}{\mathrm{2}} \\ $$$${so}\:{I}\:=\frac{\pi}{\mathrm{2}}\:. \\ $$
Answered by MJS last updated on 04/Jun/18
![∫_0 ^1 (dx/( (√(1−x^2 ))))=[arcsin x]_0 ^1 =(π/2) ...nothing improper about this...](https://www.tinkutara.com/question/Q36731.png)
$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{{dx}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}=\left[\mathrm{arcsin}\:{x}\right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{\pi}{\mathrm{2}} \\ $$$$…\mathrm{nothing}\:\mathrm{improper}\:\mathrm{about}\:\mathrm{this}… \\ $$