Question Number 93633 by abdomathmax last updated on 14/May/20 $${calvulate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left(\pi{x}\right)}{\mathrm{1}+{x}^{\mathrm{4}} }{dx} \\ $$ Commented by mathmax by abdo last updated on 15/May/20 $${A}\:=\int_{\mathrm{0}}…
Question Number 28073 by abdo imad last updated on 20/Jan/18 $${find}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−\mathrm{2}{x}} {ln}\left(\mathrm{1}+{t}\:{e}^{−{x}} \right){dx}\:\:\:{with}\:\:\mathrm{0}<{t}<\mathrm{1}\:\:. \\ $$ Commented by abdo imad last updated on 26/Jan/18…
Question Number 28071 by abdo imad last updated on 20/Jan/18 $${let}\:{give}\:\:{A}_{{p}} =\:\int_{\mathrm{0}} ^{\pi} \:{t}^{{p}} \:{cos}\left({nx}\right)\:\:{with}\:{nand}\:{p}\:{from}\:{N} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{relation}\:{between}\:\:{A}_{{p}} \:{and}\:{A}_{{p}−\mathrm{2}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{arelation}\:{between}\:\:{A}_{\mathrm{2}{p}} \:\:{and}\:{A}_{\mathrm{2}{p}−\mathrm{2}} \\ $$$$\left.\mathrm{3}\right)\:{find}\:{a}\:{relation}?{betweer}\:{A}_{\mathrm{2}{p}+\mathrm{1}} \:{and}\:\:{A}_{\mathrm{2}{p}−\mathrm{1}} \\…
Question Number 28072 by abdo imad last updated on 20/Jan/18 $${let}\:{give}\:{the}\:{function}\:\:{f}\left({x}\right)={x}^{\mathrm{4}} \:\:\:\mathrm{2}\pi\:{periodic}\:{and}\:{even} \\ $$$${developp}\:\:\:{f}\:{atfourier}\:{series}. \\ $$ Commented by abdo imad last updated on 26/Jan/18 $${f}\left(−{x}\right)={f}\left({x}\right)\:{and}\:{f}\:\mathrm{2}\pi\:{periodic}\:{so}…
Question Number 159143 by mnjuly1970 last updated on 13/Nov/21 Answered by qaz last updated on 14/Nov/21 $$\mathrm{S}=\frac{\mathrm{1}}{\Gamma\left(\mathrm{4}\right)}\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\Gamma\left(\mathrm{4n}+\mathrm{1}\right)\Gamma\left(\mathrm{4}\right)}{\Gamma\left(\mathrm{4n}+\mathrm{5}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}^{\mathrm{4n}}…
Question Number 28068 by abdo imad last updated on 19/Jan/18 $${let}\:{give}\:\:\:{I}_{{a}} \:\:=\:\:\int_{\mathrm{0}} ^{+\propto} \:\:\:\:\frac{{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:\:\:{by}\:{using}\:{Residus}\:{theorem} \\ $$$${find}\:{the}\:{value}\:{of}\:\:{I}_{{a}} \:\:\:\:\:{with}\:\:\mathrm{0}<{a}<\mathrm{1}\:\:\:. \\ $$ Terms of Service Privacy Policy…
Question Number 28067 by abdo imad last updated on 19/Jan/18 $${let}\:{give}\:{f}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{2}+{cosx}}\:\:\:{fonction}\:\mathrm{2}\pi\:{periodic}\:{even}. \\ $$$${developp}\:{f}\:\:{at}\:{fourier}\:{series}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 93587 by M±th+et+s last updated on 13/May/20 $$\int\frac{{x}+\mathrm{1}}{{x}^{\mathrm{2}} −\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}{x}+\mathrm{1}}{dx} \\ $$ Answered by niroj last updated on 13/May/20 $$\:\:\:\:\mathrm{I}=\:\int\:\frac{\:\:\mathrm{x}+\mathrm{1}}{\mathrm{x}^{\mathrm{2}} −\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\mathrm{x}+\mathrm{1}}\mathrm{dx} \\ $$$$\:=\:\int\:\frac{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2x}−\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)+\:\frac{\mathrm{5}+\sqrt{\mathrm{5}}}{\mathrm{4}}}{\mathrm{x}^{\mathrm{2}} −\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\mathrm{x}+\mathrm{1}}\mathrm{dx}…
Question Number 28041 by suraj kumar last updated on 19/Jan/18 $$\int\frac{\varkappa^{\mathrm{2}} }{\left(\varkappa\mathrm{sin}\varkappa+\mathrm{cos}\varkappa\right)^{\mathrm{2}} }\mathrm{d}\left(\varkappa\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 28038 by abdo imad last updated on 19/Jan/18 $${let}\:{give}\:{f}\left({x}\right)=\sqrt{{x}+{y}}\:+\mathrm{1}\:\:{and}\:{D}=\left\{\left({x},{y}\right)\in{R}^{\mathrm{2}} /\:\mathrm{0}\leqslant{x}\leqslant\mathrm{1}\:\right. \\ $$$$\left.{and}\:−\mathrm{1}\leqslant{y}\leqslant\mathrm{1}\right\}\:\:{find}\:{the}\:{value}\:{of}\:\:\int\int\:{f}\left({x},{y}\right){dxdy}\:. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com