Question Number 73131 by behi83417@gmail.com last updated on 06/Nov/19 $$\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{terms}}\:\boldsymbol{\mathrm{of}}:\:\:\boldsymbol{\mathrm{a}}\in\boldsymbol{\mathrm{R}}\:. \\ $$$$\:\:\:\boldsymbol{\mathrm{x}}+\sqrt{\boldsymbol{\mathrm{x}}}+\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}}+\sqrt{\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{a}}^{\mathrm{2}} }=\boldsymbol{\mathrm{a}}^{\mathrm{2}} \\ $$ Commented by mr W last updated on 06/Nov/19 $${x}\geqslant\mathrm{0}…
Question Number 138661 by mr W last updated on 16/Apr/21 Commented by mr W last updated on 16/Apr/21 $${proof}\:{for}\:{Q}\mathrm{138519} \\ $$ Commented by mr W…
Question Number 7590 by FilupSmith last updated on 05/Sep/16 $${S}\:=\:\underset{{n}=\mathrm{2}} {\overset{{k}} {\sum}}\:\frac{\mathrm{2}\left({n}+\mathrm{1}\right)}{{n}\left({n}−\mathrm{1}\right)} \\ $$$${S}=? \\ $$ Commented by Yozzia last updated on 05/Sep/16 $$\frac{\mathrm{2}\left({n}+\mathrm{1}\right)}{{n}\left({n}−\mathrm{1}\right)}\equiv\frac{{a}}{{n}}+\frac{{b}}{{n}−\mathrm{1}} \\…
Question Number 7587 by ten last updated on 05/Sep/16 $${cos}\mathrm{4}{xsin}\mathrm{4}{x} \\ $$ Answered by Rasheed Soomro last updated on 05/Sep/16 $${cos}\mathrm{4}{xsin}\mathrm{4}{x} \\ $$$$\mathrm{cos}\:\alpha\:\mathrm{sin}\:\beta=\left(\mathrm{1}/\mathrm{2}\right)\left[\mathrm{sin}\left(\alpha+\beta\right)−\mathrm{sin}\left(\alpha−\beta\right)\right]\:\: \\ $$$$\mathrm{cos}\:\mathrm{4}{x}\:\mathrm{sin}\:\mathrm{4}{x}=\left(\mathrm{1}/\mathrm{2}\right)\left[\mathrm{sin}\left(\mathrm{4}{x}+\mathrm{4}{x}\right)−\mathrm{sin}\left(\mathrm{4}{x}−\mathrm{4}{x}\right)\right]\:\:…
Question Number 138656 by ajfour last updated on 16/Apr/21 $${I}=\int\frac{{dx}}{\left({px}+{q}\right)\sqrt{{ax}^{\mathrm{2}} +{bx}+{c}}} \\ $$ Answered by Ar Brandon last updated on 16/Apr/21 $$\mathcal{I}=\int\frac{\mathrm{dx}}{\left(\mathrm{px}+\mathrm{q}\right)\sqrt{\mathrm{ax}^{\mathrm{2}} +\mathrm{bx}+\mathrm{c}}} \\ $$$$\mathrm{u}=\frac{\mathrm{1}}{\mathrm{px}+\mathrm{q}}\:\Rightarrow\mathrm{x}=\frac{\mathrm{1}}{\mathrm{up}}−\frac{\mathrm{q}}{\mathrm{p}}\Rightarrow\mathrm{du}=−\frac{\mathrm{p}}{\left(\mathrm{px}+\mathrm{q}\right)^{\mathrm{2}}…
Question Number 7585 by Tawakalitu. last updated on 04/Sep/16 $$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}}{\mathrm{1}\:+\:{x}^{\mathrm{2}} }\:{dx} \\ $$ Commented by sou1618 last updated on 05/Sep/16 $$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}}…
Question Number 138658 by ajfour last updated on 16/Apr/21 Commented by ajfour last updated on 16/Apr/21 $${Find}\:{maximum}\:{area}\:{of}\:\bigtriangleup{AOB}. \\ $$$${a}=\mathrm{1}. \\ $$ Commented by mr W…
Question Number 7582 by A WLAN last updated on 04/Sep/16 Commented by A WLAN last updated on 04/Sep/16 $${How}\:{to}\:{solve}\:{this}\:{problem}? \\ $$ Answered by Yozzia last…
Question Number 73117 by aliesam last updated on 06/Nov/19 Commented by mathmax by abdo last updated on 06/Nov/19 $${we}\:{have}\:{cos}\left(\mathrm{3}{x}\right)\sim\mathrm{1}−\frac{\left(\mathrm{3}{x}\right)^{\mathrm{2}} }{\mathrm{2}}\:\:\left({x}\rightarrow\mathrm{0}\right)\:\Rightarrow{cos}\left(\mathrm{3}{x}\right)−\mathrm{1}\:\sim−\frac{\mathrm{9}{x}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow \\ $$$$\mathrm{1}−{cos}\left(\mathrm{3}{x}\right)\sim\frac{\mathrm{9}{x}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow\frac{\mathrm{1}−{cos}\left(\mathrm{3}{x}\right)}{{x}^{\mathrm{2}} }\sim\frac{\mathrm{9}}{\mathrm{2}}\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}}…
Question Number 7579 by ajy last updated on 04/Sep/16 $$\mathrm{2}×\mathrm{2} \\ $$ Answered by FilupSmith last updated on 04/Sep/16 $$\mathrm{4} \\ $$ Terms of Service…