Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 36181 by prof Abdo imad last updated on 30/May/18

$$letI\left(\xi \right)={\int }_{\xi }^{1-\xi }\frac{dt}{1-{(t-\xi )}^2}$$$$find{lim}_{\xi \to 0^{+}}I\left(\xi \right)$$

Commented by maxmathsup by imad last updated on 20/Aug/18

$$changementt-\xi =sin\alpha give\alpha =arcsin(t-\xi )\Rightarrow$$$$I\left(\xi \right)={\int }_0^{arcsin(1-2\xi )}\frac{cos\alpha d\alpha }{1-{sin}^2\alpha }={\int }_0^{arcsin(1-2\xi )}\frac{d\alpha }{cos\alpha }$$$${=}_{tan\left(\frac{\alpha }{2}\right)=u}{\int }_0^{tan\left(\frac{arcsin(1-2\xi )}{2}\right)}\frac{1}{\frac{1-u^2}{1+u^2}}\frac{2du}{1+u^2}={\int }_0^{tan\left(\frac{arcsin(1-2\xi )}{2}\right)}\frac{2du}{1-u^2}$$$$={\int }_0^{tan\left(\frac{arcsin(1-2\xi )}{2}\right)}(\frac{1}{1+u}+\frac{1}{1-u})du={[ln\mid \frac{1+u}{1-u}\mid ]}_0^{tan\left(\frac{arcsin(1-2\xi )}{2}\right)}$$$$=ln\mid \frac{1+tan\left(\frac{arcsin(1-2\xi )}{2}\right)}{1-tan\left(\frac{arcsin(1-2\xi )}{2}\right)}\mid \Rightarrow {lim}_{\xi \to 0^{+}}I\left(\xi \right)$$$$=ln\mid \frac{1+tan\left(\frac{\pi }{4}\right)}{1-tan\left(\frac{\pi }{4}\right)}\mid =+\mathrm{\infty }.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com