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Category: Integration

Question-109658

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-109642

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-44092

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}}…

If-f-x-continue-in-1-30-and-6-30-f-x-dx-30-then-1-9-f-3y-3-dy-

Question Number 109546 by bemath last updated on 24/Aug/20 $${If}\:{f}\left({x}\right)\:{continue}\:{in}\:\left[\:\mathrm{1},\mathrm{30}\right]\:{and}\: \\ $$$$\underset{\mathrm{6}} {\overset{\mathrm{30}} {\int}}{f}\left({x}\right){dx}\:=\:\mathrm{30},\:{then}\:\underset{\mathrm{1}} {\overset{\mathrm{9}} {\int}}{f}\left(\mathrm{3}{y}+\mathrm{3}\right){dy}\:=\:\_\_ \\ $$ Commented by kaivan.ahmadi last updated on 24/Aug/20…