Category: Integration

Question Number 175244 by infinityaction last updated on 24/Aug/22 $$provethat$$$$\:\int_{\mathrm{0}} ^{{a}} \int_{\mathrm{0}} ^{\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }} \frac{{dx}\:{dy}}{\left(\mathrm{1}+{e}^{{y}} \right)\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }}\:=\:\frac{\pi}{\mathrm{2}}\mathrm{log}\:\frac{\mathrm{2}{e}^{{a}} }{\mathrm{1}+{e}^{{a}} } \…

Question Number 44174 by abdo.msup.com last updated on 22/Sep/18 $$find{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{(1+x^2)(1+x{e}^{i\theta })}$$ Terms of Service Privacy Policy Contact: info@tinkutara.com

Question Number 44173 by abdo.msup.com last updated on 22/Sep/18 $$calculate{\int }_0^{\mathrm{\infty }}\frac{dt}{(3+t^2)\sqrt{1+t}}dt$$ Commented by maxmathsup by imad last updated on 23/Sep/18 $${let}\:{I}\:=\int_{\mathrm{0}}…

Question Number 109658 by 150505R last updated on 24/Aug/20 Answered by mathmax by abdo last updated on 25/Aug/20 $$\mathrm{A}={\int }_0^{\frac{\mathrm{e}}{\pi }}\frac{\arctan \left(\frac{\pi \mathrm{x}}{\mathrm{e}}\right)}{\pi \mathrm{x}+\mathrm{e}}\mathrm{dx}\mathrm{changement}\frac{\pi \mathrm{x}}{\mathrm{e}}=\mathrm{t}\mathrm{give}$$$$\mathrm{A}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{arctan}\left(\mathrm{t}\right)}{\mathrm{et}\:+\mathrm{e}}.\frac{\mathrm{e}}{\pi}\:\mathrm{dt}\:=\frac{\mathrm{1}}{\pi}\:\int_{\mathrm{0}}…

Question Number 109642 by malwan last updated on 24/Aug/20 Answered by Dwaipayan Shikari last updated on 24/Aug/20 $$\int \frac{dx}{\sqrt{2{tan}^2\theta +2}}\frac{\sqrt{2}}{5}{sec}^2\theta d\theta x=\frac{\sqrt{2}}{5}tan\theta$$$$\frac{1}{5}\int sec\theta d\theta$$$$\frac{\mathrm{1}}{\mathrm{5}}{log}\left({sec}\theta+{tan}\theta\right)=\frac{\mathrm{1}}{\mathrm{5}}{log}\left(\sqrt{\frac{\mathrm{25}{x}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}}}\:+\frac{\mathrm{5}{x}}{\:\sqrt{\mathrm{2}}}\right)+{C}…