Question Number 121908 by rs4089 last updated on 12/Nov/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 121886 by oustmuchiya@gmail.com last updated on 12/Nov/20 Commented by TANMAY PANACEA last updated on 12/Nov/20 $${plz}\:{purchase}\:{a}\:{good}\:{text}\:{book}\:{and}\:{study}\:{the} \\ $$$${basic} \\ $$ Answered by mathmax…
Question Number 121887 by oustmuchiya@gmail.com last updated on 12/Nov/20 Commented by liberty last updated on 12/Nov/20 $$\left(\mathrm{vi}\right)\:\left(\mathrm{2cos}\:\mathrm{x}−\mathrm{sin}\:\mathrm{x}\right)^{\mathrm{2}} =\mathrm{4cos}\:^{\mathrm{2}} \mathrm{x}−\mathrm{4cos}\:\mathrm{xsin}\:\mathrm{x}+\mathrm{sin}\:^{\mathrm{2}} \mathrm{x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{4}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2x}\right)−\mathrm{2sin}\:\mathrm{2x}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2}\:+\:\mathrm{2cos}\:\mathrm{2x}−\mathrm{2sin}\:\mathrm{2x}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2x} \\…
Question Number 56345 by maxmathsup by imad last updated on 14/Mar/19 $${let}\:\:{f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{dx}}{{x}^{{n}} \:+{a}^{{n}} }\:\:\:{with}\:{n}\:{integr}\:\geqslant\mathrm{2}\:\:{and}\:{a}>\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}\left({a}\right)\:{intems}\:{of}\:{a} \\ $$$$\left.\mathrm{2}\right)\:{let}\:{g}\left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left({x}^{{n}} \:+{a}^{{n}} \right)^{\mathrm{2}} }\:\:{calculate}\:{g}\left({a}\right)\:{interms}\:{of}\:{a}…
Question Number 121875 by rs4089 last updated on 12/Nov/20 Answered by Dwaipayan Shikari last updated on 12/Nov/20 $$\frac{\pi^{\mathrm{4}} }{\mathrm{90}} \\ $$ Answered by Dwaipayan Shikari…
Question Number 121862 by Bird last updated on 12/Nov/20 $$\left.\mathrm{1}\right){explicite}\:{f}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{t}^{{a}−\mathrm{1}} {lnt}}{\mathrm{1}+{t}}{dt} \\ $$$${with}\:\mathrm{0}<{a}<\mathrm{1} \\ $$$$\left.\mathrm{2}\right){calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{lnt}}{\left(\mathrm{1}+{t}\right)\sqrt{{t}}}{dt} \\ $$ Answered by mnjuly1970 last…
Question Number 56329 by maxmathsup by imad last updated on 14/Mar/19 $$\left.\mathrm{1}\right){calculate}\:{A}_{{n}} =\int_{\frac{\mathrm{1}}{{n}}} ^{\mathrm{1}} \:\:\frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:\:\:\:{with}\:{n}\:{integr}\:{and}\:{n}\geqslant\mathrm{1} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{lim}_{{n}\rightarrow+\infty} \:\:\:\:{A}_{{n}} \\ $$$$\left.\mathrm{3}\right)\:\:{study}\:{the}\:{convergence}\:{of}\:\Sigma\:{A}_{{n}} \\ $$ Commented…
Question Number 121860 by Bird last updated on 12/Nov/20 $${calculate}\:\int\int_{\left[\mathrm{0},\mathrm{1}\right]^{\mathrm{2}} } \:\:\:\frac{{arctan}\left(\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\right)}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{dxdy} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 121858 by Bird last updated on 12/Nov/20 $${decompose}\:{F}\left({x}\right)\:=\frac{{x}^{{n}} }{{x}^{\mathrm{2}{n}+\mathrm{1}} +\mathrm{1}} \\ $$$$\left.\mathrm{2}\right){find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{x}^{{n}} }{{x}^{\mathrm{2}{n}+\mathrm{1}} +\mathrm{1}}{dx} \\ $$$${n}\:{integr}\:{natural} \\ $$ Terms of Service…
Question Number 121859 by Bird last updated on 12/Nov/20 $${find}\:\int\:\:\:\frac{{dx}}{\left({x}−\mathrm{1}\right)\sqrt{{x}+\mathrm{1}}−\left({x}+\mathrm{1}\right)\sqrt{{x}−\mathrm{1}}} \\ $$ Answered by ajfour last updated on 12/Nov/20 $${I}=\int\frac{\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}}{\:\sqrt{{x}−\mathrm{1}}−\sqrt{{x}+\mathrm{1}}} \\ $$$$ \\ $$$$\:\:=\int\:\frac{\left(\sqrt{{x}−\mathrm{1}}+\sqrt{{x}+\mathrm{1}}\right){dx}}{\mathrm{2}\sqrt{{x}^{\mathrm{2}}…