Question Number 210572 by Spillover last updated on 12/Aug/24 Answered by mathmax last updated on 13/Aug/24 $${I}=\int_{\mathrm{0}} ^{\infty} \:\frac{{ln}^{\mathrm{2}} {x}}{\mathrm{1}+{x}^{\mathrm{4}} }{dx}\:\:{changement}\:{x}={t}^{\frac{\mathrm{1}}{\mathrm{4}}} {give} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{16}}\int_{\mathrm{0}} ^{\infty}…
Question Number 210573 by Spillover last updated on 12/Aug/24 Answered by mathmax last updated on 13/Aug/24 $$\mathrm{2}{I}=\int_{−\infty} ^{+\infty} \frac{{dx}}{{x}^{\mathrm{4}} +{ix}^{\mathrm{2}} +\mathrm{2}}\:\left({fonction}\:{paire}\right) \\ $$$${roots}\:\:\:\:\:\:{x}^{\mathrm{2}} ={t}\rightarrow{t}^{\mathrm{2}} +{it}+\mathrm{2}…
Question Number 210542 by SANOGO last updated on 12/Aug/24 Commented by Rasheed.Sindhi last updated on 12/Aug/24 $${x}+{y}+{z}=\mathrm{1}\Rightarrow\left({x}+{y}+{z}\right)^{\mathrm{2}} =\mathrm{1} \\ $$ Commented by Frix last updated…
Question Number 210571 by hardmath last updated on 12/Aug/24 $$\mathrm{If}\:\:\mathrm{x},\mathrm{y},\mathrm{z}\in\mathrm{R}^{+} \:\:\mathrm{and}\:\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} =\mathrm{3} \\ $$$$\mathrm{Prove}\:\mathrm{that} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}−\mathrm{x}}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{4}−\mathrm{y}}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{4}−\mathrm{z}}\:\:\leqslant\:\:\mathrm{1} \\ $$ Answered by A5T last updated…
Question Number 210565 by maths_plus last updated on 12/Aug/24 Answered by mr W last updated on 13/Aug/24 Commented by mr W last updated on 13/Aug/24…
Question Number 210566 by Erico last updated on 12/Aug/24 $$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\mathrm{if}\:\left(\mathrm{x}\in\right]−\frac{\pi}{\mathrm{2}},\frac{\pi}{\mathrm{2}}\left[\:\:\mathrm{y}\:=\underset{\:\mathrm{0}} {\int}^{\:\mathrm{x}} \frac{\mathrm{dt}}{\mathrm{cos}\left(\mathrm{t}\right)}\:\right)\:\Rightarrow\:\:\left(\mathrm{y}\in\mathrm{IR}\:\:\:\mathrm{x}\:=\underset{\:\mathrm{0}} {\int}^{\:\mathrm{y}} \frac{\mathrm{dt}}{\mathrm{cosh}\left(\mathrm{t}\right)}\:\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 210524 by hardmath last updated on 11/Aug/24 Commented by hardmath last updated on 11/Aug/24 $$\boldsymbol{\mathrm{r}}\:=\:? \\ $$ Answered by mr W last updated…
Question Number 210492 by a.lgnaoui last updated on 11/Aug/24 $$\mathrm{un}\:\mathrm{objet}\:\boldsymbol{\mathrm{C}}\:\mathrm{situe}\:\mathrm{a}\:\mathrm{une}\:\mathrm{hsuteur}\:\boldsymbol{\mathrm{h}} \\ $$$$\mathrm{est}\:\mathrm{lache}\:\mathrm{sans}\:\mathrm{vitesse}\:\:\mathrm{initiale}\:\left(\boldsymbol{\mathrm{v}}_{\mathrm{0}} =\mathrm{0}\right) \\ $$$$\mathrm{par}\:\mathrm{un}\:\mathrm{un}\:\mathrm{petit}\:\mathrm{avion}\:\mathrm{volant}\:\mathrm{a}\:\mathrm{basse} \\ $$$$\mathrm{altitue}\:\mathrm{cet}\:\mathrm{objet}\:\mathrm{est}\:\mathrm{capte}\:\mathrm{par}\:\mathrm{une}\: \\ $$$$\mathrm{prrsonne}\:\:\mathrm{entraine}\:\mathrm{le}\:\mathrm{long}\:\mathrm{de}\:\mathrm{la}\:\:\mathrm{couronne}\: \\ $$$$\mathrm{a}\:\mathrm{vitesse}\:\mathrm{const}\boldsymbol{\mathrm{a}}\mathrm{nte}\left(\boldsymbol{\mathrm{v}}\right)\: \\ $$$$\left.−\mathrm{1}\right)\:\mathrm{Determiner}\:\mathrm{l}\:\mathrm{instant}\:\mathrm{de}\:\mathrm{recuperariion} \\ $$$$\mathrm{de}\:\mathrm{l}\:\mathrm{objet}\:…
Question Number 210516 by hardmath last updated on 11/Aug/24 $$\mathrm{Find}:\:\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$$$\mathrm{1}\:+\:\mathrm{3x}\:+\:\mathrm{5x}^{\mathrm{2}} \:+\:\mathrm{7x}^{\mathrm{3}} \:+\:\mathrm{9x}^{\mathrm{4}} \:+\:\mathrm{11x}^{\mathrm{5}} \:+\:…\:=\:\mathrm{15} \\ $$ Answered by mm1342 last updated on 12/Aug/24…
Question Number 210517 by depressiveshrek last updated on 11/Aug/24 $$\int\frac{\mathrm{1}}{\mathrm{sin}{x}−\mathrm{cos2}{x}}{dx} \\ $$ Commented by Frix last updated on 11/Aug/24 $$\mathrm{Simply}\:\mathrm{use}\:{t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:\mathrm{to}\:\mathrm{get} \\ $$$$\frac{\mathrm{2}}{\mathrm{3}\left(\mathrm{1}+\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\right)}+\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{9}}\mathrm{ln}\:\mid\frac{\mathrm{2}−\sqrt{\mathrm{3}}−\mathrm{tan}\:\frac{{x}}{\mathrm{2}}}{\mathrm{2}+\sqrt{\mathrm{3}}−\mathrm{tan}\:\frac{{x}}{\mathrm{3}}}\mid\:+{C} \\ $$ Commented…