Question Number 148720 by qaz last updated on 30/Jul/21 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{Talor}\:\mathrm{series}\:\mathrm{of}\:\frac{\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)}{\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{2}} }\:\mathrm{at}\:\mathrm{x}=\mathrm{0}. \\ $$ Answered by mathmax by abdo last updated on 30/Jul/21 $$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)}{\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{2}} }\:\:\mathrm{we}\:\mathrm{have}\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty}…
Question Number 17625 by tawa tawa last updated on 08/Jul/17 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{fourier}\:\mathrm{series}\:\mathrm{of}\::\:\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{x},\:\:\mathrm{from}\:\:\:\mathrm{0}\:<\:\mathrm{x}\:<\:\pi \\ $$ Commented by tawa tawa last updated on 08/Jul/17 $$\mathrm{please}\:\mathrm{help}\:\mathrm{with}\:\mathrm{this}. \\ $$ Answered…
Question Number 83164 by niroj last updated on 28/Feb/20 $$ \\ $$$$ \\ $$$$\:\mathrm{Evaluate}: \\ $$$$\:\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \:\frac{\:\:\mathrm{1}}{\mathrm{1}+\boldsymbol{\mathrm{cos}}\:\boldsymbol{\alpha}\:\boldsymbol{\mathrm{cos}}\:\boldsymbol{\mathrm{x}}}\boldsymbol{\mathrm{dx}} \\ $$ Commented by mathmax by abdo…
Question Number 17604 by tawa tawa last updated on 08/Jul/17 $$\int_{\:\:\mathrm{0}} ^{\:\:\mathrm{n}} \:\mathrm{x}^{\mathrm{2}} \left(\mathrm{n}\:−\:\mathrm{x}\right)^{\mathrm{p}} \:\mathrm{dx}\:\:\:\:\:\:\:\mathrm{for}\:\:\:\mathrm{p}\:>\:\mathrm{0} \\ $$ Answered by sma3l2996 last updated on 08/Jul/17 $${t}={n}−{x}\Rightarrow{dt}=−{dx}…
Question Number 17603 by tawa tawa last updated on 08/Jul/17 $$\int_{\:\:\mathrm{0}} ^{\:\:\mathrm{a}/\mathrm{2}} \:\mathrm{x}^{\mathrm{2}} \left(\mathrm{a}^{\mathrm{2}} \:−\:\mathrm{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:\mathrm{dx} \\ $$ Answered by alex041103 last updated on…
Question Number 17599 by chux last updated on 08/Jul/17 $$\mathrm{a}\:\mathrm{particle}\:\mathrm{starts}\:\mathrm{with}\:\mathrm{an}\:\mathrm{initial} \\ $$$$\mathrm{speed}\:\mathrm{u},\mathrm{it}\:\mathrm{moves}\:\mathrm{in}\:\mathrm{a}\:\mathrm{straight} \\ $$$$\mathrm{line}\:\mathrm{with}\:\mathrm{an}\:\mathrm{accleration}\:\mathrm{which} \\ $$$$\mathrm{varies}\:\mathrm{as}\:\mathrm{the}\:\mathrm{square}\:\mathrm{of}\:\mathrm{the}\:\mathrm{time} \\ $$$$\mathrm{the}\:\mathrm{particle}\:\mathrm{has}\:\mathrm{been}\:\mathrm{in}\:\mathrm{motion}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{speed}\:\mathrm{at}\:\mathrm{any}\:\mathrm{time}\:\mathrm{t},\mathrm{and} \\ $$$$\mathrm{the}\:\mathrm{distance}\:\mathrm{travelled}. \\ $$ Answered…
Question Number 83123 by M±th+et£s last updated on 28/Feb/20 $$\int\frac{\left({x}^{\mathrm{2}} −\mathrm{1}\right)}{\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right)\left({x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{2}\right)}\:{dx} \\ $$ Commented by mathmax by abdo last updated on 28/Feb/20 $${A}\:=\int\:\:\frac{{x}^{\mathrm{2}}…
Question Number 83115 by 09658867628 last updated on 28/Feb/20 $$\int\mathrm{cos}\:{xe}^{\mathrm{sin}\:{x}} {dx} \\ $$ Commented by niroj last updated on 28/Feb/20 $$ \\ $$$$\:\int\boldsymbol{\mathrm{cos}}\:\boldsymbol{\mathrm{x}}\:\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{sin}}\:\boldsymbol{\mathrm{x}}} \boldsymbol{\mathrm{dx}} \\…
Question Number 83108 by M±th+et£s last updated on 28/Feb/20 $${prove}\:{that} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{cos}\left({nx}\right)}{{cos}^{{n}} \left({x}\right)}\:{dx}\:=\mathrm{2}^{{n}} \left[\frac{\pi}{\mathrm{8}}−\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{{sin}\left(\frac{{k}\pi}{\mathrm{4}}\right)}{\mathrm{2}{k}\left(\sqrt{\mathrm{2}}\right)^{{k}} }\right]\:{n}\in{N}^{\ast} \\ $$ Answered by mind is…
Question Number 83110 by 09658867628 last updated on 28/Feb/20 $${bounded}\:{by}\:{the}\:{curve}\:{y}=\sqrt{\mathrm{4}-{x}}\:{y}=\mathrm{0}\:{y}=\mathrm{1} \\ $$ Commented by jagoll last updated on 28/Feb/20 $$\mathrm{Area}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{4}−\mathrm{y}^{\mathrm{2}} \right)\:\mathrm{dy}\: \\ $$$$=\:\mathrm{4y}\:−\:\frac{\mathrm{y}^{\mathrm{3}}…