Question Number 175244 by infinityaction last updated on 24/Aug/22 $$\:{prove}\:{that} \\ $$$$\:\int_{\mathrm{0}} ^{{a}} \int_{\mathrm{0}} ^{\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }} \frac{{dx}\:{dy}}{\left(\mathrm{1}+{e}^{{y}} \right)\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }}\:=\:\frac{\pi}{\mathrm{2}}\mathrm{log}\:\frac{\mathrm{2}{e}^{{a}} }{\mathrm{1}+{e}^{{a}} } \\…
Question Number 175247 by rexford last updated on 24/Aug/22 Answered by Ar Brandon last updated on 24/Aug/22 $${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\mathrm{2}} \mathrm{4}\vartheta\mathrm{cos}^{\mathrm{5}} \mathrm{4}\vartheta{d}\vartheta \\ $$$$\:\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}}…
Question Number 44174 by abdo.msup.com last updated on 22/Sep/18 $${find}\:\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}\:{e}^{{i}\theta} \right)} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 109709 by nimnim last updated on 25/Aug/20 $$\:\:\:\:\:\int_{\mathrm{0}\:} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{tanx}+\mathrm{1}\right)\mathrm{dx} \\ $$ Answered by mnjuly1970 last updated on 25/Aug/20 $$\mathrm{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left(\mathrm{1}+{tanx}\right){dx}\overset{\int_{{a}} ^{\:{b}}…
Question Number 44173 by abdo.msup.com last updated on 22/Sep/18 $${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dt}}{\left(\mathrm{3}+{t}^{\mathrm{2}} \right)\sqrt{\mathrm{1}+{t}}}{dt} \\ $$ Commented by maxmathsup by imad last updated on 23/Sep/18 $${let}\:{I}\:=\int_{\mathrm{0}}…
Question Number 44148 by LXZ last updated on 22/Sep/18 $$\int{dx}/{sinx}\centerdot{sin}\left({x}+\alpha\right)=? \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 22/Sep/18 $$\int\frac{{dx}}{{sinx}.{sin}\left({x}+\alpha\right)} \\ $$$$\frac{\mathrm{1}}{{sin}\alpha}\int\frac{{sin}\left\{\left({x}+\alpha\right)−{x}\right\}}{{sinx}.{sin}\left({x}+\alpha\right)}{dx} \\ $$$$\frac{\mathrm{1}}{{sin}\alpha}\int\frac{{sin}\left({x}+\alpha\right){cosx}−{cos}\left({x}+\alpha\right){sinx}}{{sinxsin}\left({x}+\alpha\right)}{dx} \\…
Question Number 109658 by 150505R last updated on 24/Aug/20 Answered by mathmax by abdo last updated on 25/Aug/20 $$\mathrm{A}\:=\int_{\mathrm{0}} ^{\frac{\mathrm{e}}{\pi}} \:\frac{\mathrm{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} \\ $$$$\mathrm{A}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{arctan}\left(\mathrm{t}\right)}{\mathrm{et}\:+\mathrm{e}}.\frac{\mathrm{e}}{\pi}\:\mathrm{dt}\:=\frac{\mathrm{1}}{\pi}\:\int_{\mathrm{0}}…
Question Number 175191 by ajfour last updated on 22/Aug/22 $$\int\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{x}}}{dx}=? \\ $$ Commented by ajfour last updated on 22/Aug/22 $${Thank}\:{you}\:{both}\:{sirs}. \\ $$ Answered by Ar…
Question Number 109642 by malwan last updated on 24/Aug/20 Answered by Dwaipayan Shikari last updated on 24/Aug/20 $$\int\frac{{dx}}{\:\sqrt{\mathrm{2}{tan}^{\mathrm{2}} \theta+\mathrm{2}}}\:\frac{\sqrt{\mathrm{2}}}{\mathrm{5}}{sec}^{\mathrm{2}} \theta{d}\theta\:\:\:\:\:\:\:\:\:\:{x}=\frac{\sqrt{\mathrm{2}}}{\mathrm{5}}{tan}\theta \\ $$$$\frac{\mathrm{1}}{\mathrm{5}}\int{sec}\theta{d}\theta \\ $$$$\frac{\mathrm{1}}{\mathrm{5}}{log}\left({sec}\theta+{tan}\theta\right)=\frac{\mathrm{1}}{\mathrm{5}}{log}\left(\sqrt{\frac{\mathrm{25}{x}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}}}\:+\frac{\mathrm{5}{x}}{\:\sqrt{\mathrm{2}}}\right)+{C}…
Question Number 44092 by peter frank last updated on 21/Sep/18 Commented by maxmathsup by imad last updated on 21/Sep/18 $${let}\:{I}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)}\:\:{changement}\:{x}\:={sh}\left({t}\right)\:{give}\: \\ $$$${I}\:=\:\int_{\mathrm{0}}…