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 119306 by yahyajan last updated on 23/Oct/20

Answered by TANMAY PANACEA last updated on 23/Oct/20

$$x=dtana$$$${\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}\frac{{dsec}^{2}da}{d^2{sec}^{2}a}$$$$\frac{1}{d}(\frac{\pi }{2}+\frac{\pi }{2})=\frac{\pi }{d}$$

Answered by Olaf last updated on 23/Oct/20

$$\mathrm{I}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{x^2+d^2}={\left[\frac{1}{d}\arctan \left(\frac{x}{d}\right)\right]}_{-\mathrm{\infty }}^{+\mathrm{\infty }}$$$$\mathrm{I}=\frac{\pi }{2d}-(-\frac{\pi }{2d})=\frac{\pi }{d}$$$$$$

Answered by mathmax by abdo last updated on 24/Oct/20

$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{dx}}{{\mathrm{x}}^2+{\mathrm{d}}^2}{=}_{\mathrm{x}=\mid \mathrm{d}\mid \mathrm{u}}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mid \mathrm{d}\mid \mathrm{du}}{{\mathrm{d}}^2(1+{\mathrm{u}}^2)}=\frac{1}{\mid \mathrm{d}\mid }{\left[\mathrm{arctanu}\right]}_{-\mathrm{\infty }}^{+\mathrm{\infty }}=\frac{\pi }{\mid \mathrm{d}\mid }$$$$(\mathrm{if}\mathrm{d}\ne 0)\mathrm{if}\mathrm{d}=0\mathrm{we}\mathrm{get}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{dx}}{{\mathrm{x}}^2}={[-\frac{1}{\mathrm{x}}]}_{-\mathrm{\infty }}^{+\mathrm{\infty }}=0$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com