Question Number 47101 by Meritguide1234 last updated on 04/Nov/18 Commented by tanmay.chaudhury50@gmail.com last updated on 05/Nov/18 Answered by tanmay.chaudhury50@gmail.com last updated on 05/Nov/18 $$\int\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}}…
Question Number 47098 by rahul 19 last updated on 04/Nov/18 $${prove}\:{that}:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\int_{−\mathrm{1}} ^{\mathrm{1}} \left(\mathrm{1}+\frac{{t}}{{n}}\right)^{{n}} {dt}\:=\:{e}−\frac{\mathrm{1}}{{e}}. \\ $$ Commented by rahul 19 last updated on 04/Nov/18…
Question Number 47088 by Meritguide1234 last updated on 04/Nov/18 Answered by MJS last updated on 04/Nov/18 $${x}^{\mathrm{2}} +{ax}+{b}=\mathrm{0} \\ $$$${x}=−\frac{{a}}{\mathrm{2}}\pm\frac{\sqrt{{a}^{\mathrm{2}} −\mathrm{4}{b}}}{\mathrm{2}}\:\Rightarrow\:\alpha=−\frac{{a}}{\mathrm{2}}−\frac{\sqrt{{a}^{\mathrm{2}} −\mathrm{4}{b}}}{\mathrm{2}};\:\beta=−\frac{{a}}{\mathrm{2}}+\frac{\sqrt{{a}^{\mathrm{2}} −\mathrm{4}{b}}}{\mathrm{2}\left(\right.} \\ $$$${bx}^{\mathrm{2}}…
Question Number 112613 by mnjuly1970 last updated on 08/Sep/20 $$\:\:\:\:\:….{number}\:{theory}… \\ $$$$\:\:\:\:\:\:\:{Question}\::\:\:\:\:\:\:\mathrm{I}{f}\:\:\:{a}\:,\:{b}\:,\:{c}\:\:\in\:\mathbb{N}\:\:\:;\:{then}\: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{prove}\:::: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}!\ast{b}!\ast{c}!\mid\left({a}+{b}+{c}\right)!\:\: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{m}.{n}\:.\:{july}\:\mathrm{970}# \\…
Question Number 47065 by maxmathsup by imad last updated on 04/Nov/18 $${let}\:{v}_{{n}} \left({a}\right)=\:\int_{\frac{\mathrm{1}}{{n}}} ^{{n}} \:\:\left(\mathrm{1}−\frac{{a}}{{x}^{\mathrm{2}} }\right){arctan}\left(\mathrm{1}+\frac{{a}}{{x}}\right){dx}\:\:{with}\:{a}>\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{determine}\:{a}\:{explicit}\:{form}\:{of}\:{v}_{{n}} \left({a}\right) \\ $$$$\left.\mathrm{2}\right)\:{study}\:{the}\:{convergence}\:{of}\:\sum_{{n}} \:{v}_{{n}} \left({a}\right) \\ $$$$\left.\mathrm{3}\right){calculate}\:{v}_{{n}}…
Question Number 47062 by maxmathsup by imad last updated on 04/Nov/18 $${find}\:\int\:\:\:\sqrt{\frac{\sqrt{{x}}−\mathrm{1}}{\:\sqrt{{x}}+\mathrm{1}}}{dx} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 04/Nov/18 $${t}^{\mathrm{2}} ={x}\:\:\:{dx}=\mathrm{2}{tdt} \\ $$$$\int\sqrt{\frac{{t}−\mathrm{1}}{{t}+\mathrm{1}}}\:×\mathrm{2}{tdt}…
Question Number 47061 by maxmathsup by imad last updated on 04/Nov/18 $${let}\:\:{f}\left({a}\right)\:=\int\:\:\:\sqrt{\mathrm{1}+{atan}\left({x}\right)}{dx} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{explicit}\:{form}\:{of}\:{f}\left({a}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int\:\:\sqrt{\mathrm{1}+\mathrm{2}{tan}\left({x}\right)}{dx}\:. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 47059 by maxmathsup by imad last updated on 04/Nov/18 $$\:{find}\:\int\:\:\frac{{dx}}{\mathrm{1}+{cos}\:{x}\:+{cos}\left(\mathrm{2}{x}\right)} \\ $$ Commented by maxmathsup by imad last updated on 06/Nov/18 $${let}\:{A}\:=\int\:\:\:\frac{{dx}}{\mathrm{1}+{cosx}\:+{cos}\left(\mathrm{2}{x}\right)}\:=\:\int\:\:\frac{{dx}}{\mathrm{1}+{cos}\left({x}\right)+\mathrm{2}{cos}^{\mathrm{2}} {x}−\mathrm{1}}…
Question Number 47060 by maxmathsup by imad last updated on 04/Nov/18 $${let}\:{f}\left({x}\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{dt}}{\mathrm{2}+{ch}\left({xt}\right)} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{explicit}\:{form}\:{of}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{g}\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{tsh}\left({xt}\right)}{\left(\mathrm{2}+{ch}\left({xt}\right)\right)^{\mathrm{2}} }{dt} \\ $$$$\left.\mathrm{3}\right)\:{find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{\mathrm{2}+{ch}\left(\mathrm{3}{t}\right)}\:{and}\:\:\int_{\mathrm{0}}…
Question Number 47058 by maxmathsup by imad last updated on 04/Nov/18 $${calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{1}+{x}^{\mathrm{3}} }{\mathrm{1}+{x}^{\mathrm{4}} }{dx} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 04/Nov/18…