Question Number 167526 by Gbenga last updated on 18/Mar/22 $$\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\boldsymbol{\sum}}}\frac{\mathrm{1}}{\boldsymbol{\mathrm{n}}^{\mathrm{2}} +\mathrm{1}}\boldsymbol{\mathrm{dn}}=??? \\ $$ Commented by aleks041103 last updated on 18/Mar/22 $${how}\:{do}\:{you}\:{sum}\:{and}\:{integrate}\:{over}…
Question Number 167517 by Gbenga last updated on 18/Mar/22 $$\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\boldsymbol{\sum}}}\frac{\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\mathrm{n}}\right)\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{n}}\right)}{\boldsymbol{\mathrm{tan}}\left(\boldsymbol{\mathrm{n}}\right)} \\ $$ Answered by alephzero last updated on 18/Mar/22 $$\frac{\mathrm{cos}\:{n}\:\mathrm{sin}\:{n}}{\mathrm{tan}\:{n}}\:=\:\frac{\mathrm{cos}\:{n}\:\mathrm{sin}\:{n}}{\frac{\mathrm{sin}\:{n}}{\mathrm{cos}\:{n}}}\:= \\ $$$$=\:\frac{\mathrm{cos}\:{n}\:\mathrm{sin}\:{n}\:\mathrm{cos}\:{n}}{\mathrm{sin}\:{n}}\:=\:\mathrm{cos}^{\mathrm{2}} \:{n}…
Question Number 167508 by Bagus1003 last updated on 18/Mar/22 $$\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{2}{x}+\mathrm{1}}={x}^{\mathrm{3}} −\mathrm{1} \\ $$$${How}\:{much}\:{the}\:{x}\:{is}? \\ $$ Answered by alephzero last updated on 18/Mar/22 $$\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{2}{x}+\mathrm{1}}\:=\:{x}^{\mathrm{3}} −\mathrm{1} \\…
Question Number 167505 by Bagus1003 last updated on 18/Mar/22 $${Solve}\:{this}\:{Equation}\:: \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\left(\left(\frac{{x}}{−{x}}\right)×\left(\left(\frac{{x}}{{x}}\right)+\left(\frac{{x}}{{x}}\right)\right)^{\left(\left(\frac{{x}}{{x}}\right)+\left(\frac{{x}}{{x}}\right)+\left(\frac{{x}}{{x}}\right)\right)} \right) \\ $$$${Then}\:{says}\:{it}\:{in}\:{English}\:{word},\: \\ $$$${but}\:{without}\:{saying}\:{the}\:{word}\: \\ $$$$''\boldsymbol{{N}}{egative}'' \\ $$$${That}\:{would}\:{be}\:{NSFW}\:{account}\:{on} \\ $$$${Twitter}. \\…
Question Number 167493 by mathlove last updated on 18/Mar/22 $$\sqrt{{x}+\mathrm{6}}+\sqrt{\mathrm{8}−{x}}={A} \\ $$$${x}\in{Z}\:\:\:{and}\:{A}\in{R}\:\:\:\:\:\:\:\:\:\overset{\:\:\:\:{faid}\:\:\Sigma{x}=?} {\:} \\ $$ Answered by Rasheed.Sindhi last updated on 18/Mar/22 $$\sqrt{{x}+\mathrm{6}}+\sqrt{\mathrm{8}−{x}}={A} \\ $$$${x}\in{Z}\:\:\:{and}\:{A}\in{R}\:\:\:\:\:\:\:\:\:{find}\:\Sigma{x}.…
Question Number 167492 by mathlove last updated on 18/Mar/22 $${x}^{{a}} =\sqrt{\mathrm{2}}+\mathrm{1}\:\:\:\:\:………\left(\mathrm{1}\right)\: \\ $$$${x}^{{b}} =\sqrt{\mathrm{2}}−\mathrm{1}\:\:\:\:\:………\left(\mathrm{2}\right) \\ $$$$\frac{\mathrm{1}}{{x}^{{a}−{b}} }+\frac{\mathrm{1}}{{x}^{{b}−{a}} }=? \\ $$$$\left(\mathrm{1}\right)\boldsymbol{\div}\left(\mathrm{2}\right)\Rightarrow\frac{{x}^{{a}} }{{x}^{{b}} }=\frac{\sqrt{\mathrm{2}}+\mathrm{1}}{\:\sqrt{\mathrm{2}−\mathrm{1}}}\Rightarrow{x}^{{a}−{b}} =\frac{\sqrt{\mathrm{2}}+\mathrm{1}}{\:\sqrt{\mathrm{2}−\mathrm{1}}}\Rightarrow\frac{\mathrm{1}}{{x}^{{a}−{b}} }=\frac{\sqrt{\mathrm{2}}−\mathrm{1}}{\:\sqrt{\mathrm{2}}+\mathrm{1}}….\left(\mathrm{3}\right) \\…
Question Number 167490 by mathlove last updated on 18/Mar/22 $$\int{sin}^{\mathrm{3}} {xcos}^{\mathrm{2}} {xdx}=? \\ $$ Answered by nimnim last updated on 18/Mar/22 $${I}=\int{sin}^{\mathrm{2}} {cos}^{\mathrm{2}} {xsinxdx} \\…
Question Number 167485 by Gbenga last updated on 17/Mar/22 $$\int\int\int\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{n}}} \boldsymbol{\mathrm{dx}} \\ $$ Answered by puissant last updated on 18/Mar/22 $$\int\int\int{x}^{{n}} {dx}\:=\:\int\int\left\{\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}+{C}\right\}{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\int\left\{\frac{{x}^{{n}+\mathrm{2}}…
Question Number 167484 by Gbenga last updated on 17/Mar/22 $$\int_{\mathrm{0}} ^{\mathrm{1}} \boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{cos}}^{−\mathrm{1}} \left(\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{dx}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 167482 by mathman1234 last updated on 17/Mar/22 $$\mathrm{montrer}\:\mathrm{que}\:\forall\mathrm{x}\in\mathbb{R}−\left\{\mathrm{0};\:\mathrm{1}\right\}\:\mathrm{on}\:\mathrm{a} \\ $$$$\frac{\mathrm{ln}\left(\mathrm{x}\right)}{\mathrm{x}−\mathrm{1}}\:<\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}}}\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com