Question Number 211932 by a.lgnaoui last updated on 25/Sep/24 $$\mathrm{determiner}\:\:\:\:\boldsymbol{\mathrm{R}}\mathrm{1}\:\:\:\:\boldsymbol{\mathrm{R}}\mathrm{2}\:\mathrm{et}\:\boldsymbol{\mathrm{R}}\mathrm{3} \\ $$$$\mathrm{segment}\:\mathrm{de}\:\mathrm{longueur}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{est}}\:\boldsymbol{\mathrm{tangent}}\:\boldsymbol{\mathrm{aux}} \\ $$$$\boldsymbol{\mathrm{cercles}}\:\mathrm{1}\boldsymbol{\mathrm{et}}\:\mathrm{2}. \\ $$$$\:\boldsymbol{\mathrm{MN}}//\boldsymbol{\mathrm{EF}};\:\:\boldsymbol{\mathrm{EF}}=\boldsymbol{\mathrm{a}};\:\:\mathrm{OM}=\mathrm{ON}=\frac{\mathrm{3}\boldsymbol{\mathrm{a}}}{\mathrm{2}}. \\ $$$$\left(\mathrm{length}\:\boldsymbol{\mathrm{a}}\:\mathrm{is}\:\mathrm{tangent}\:\mathrm{to}\:\mathrm{cirles}\:\mathrm{C1}\:\left(\mathrm{radius}\:\mathrm{R1}\right)\mathrm{and}\:\mathrm{circldC2}\left(\mathrm{radius}\:\mathrm{R2}\right)\right). \\ $$ Commented by a.lgnaoui last updated…
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Question Number 211917 by Spillover last updated on 24/Sep/24 Answered by BHOOPENDRA last updated on 24/Sep/24 $${Use}\:{L}'{Hopital}'{s}\:{Rule}\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{mn}\left(\mathrm{1}+{nx}\right)^{{m}−\mathrm{1}} −{mn}\left(\mathrm{1}+{mx}\right)^{{n}−\mathrm{1}} }{\left(\mathrm{1}+{mx}\right)^{\frac{\mathrm{1}}{{m}}−\mathrm{1}} −\left(\mathrm{1}+{nx}\right)^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} } \\…
Question Number 211919 by Spillover last updated on 24/Sep/24 Answered by Frix last updated on 24/Sep/24 $${x}^{\mathrm{1}+\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{4}}{\mathrm{4}}+\frac{\mathrm{5}}{\mathrm{8}}+\frac{\mathrm{6}}{\mathrm{16}}…} ={x}^{\mathrm{5}} \\ $$$$\mathrm{1}+\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{{k}+\mathrm{2}}{\mathrm{2}^{{k}} }\:=\mathrm{1}+\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{2}^{{k}}…
Question Number 211896 by Spillover last updated on 23/Sep/24 Answered by BHOOPENDRA last updated on 23/Sep/24 $$=−\frac{\mathrm{1}}{\mathrm{2}}\left({Re}\underset{{z}=\mathrm{0}} {{s}}\left({f}\left({z}\right)\right)+{Res}\underset{{z}=−\mathrm{1}} {\:}\:\left({f}\left({z}\right)\right)\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{n}−\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{2}{n}^{\mathrm{2}} +\mathrm{1}\right)\right) \\ $$$$=\frac{\mathrm{2}{n}^{\mathrm{2}} −\mathrm{3}{n}+\mathrm{1}}{\mathrm{3}}…
Question Number 211886 by Spillover last updated on 23/Sep/24 Answered by Ghisom last updated on 23/Sep/24 $$\int\sqrt{\mathrm{cos}^{\mathrm{2}} \:{x}\:+\mathrm{cos}\:{x}}\:{dx}=\int\sqrt{\mathrm{2cos}\:{x}}\:\mathrm{cos}\:\frac{{x}}{\mathrm{2}}\:{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{\sqrt{\mathrm{2}}\mathrm{sin}\:\frac{{x}}{\mathrm{2}}}{\:\sqrt{\mathrm{cos}\:{x}}}\:\Leftrightarrow\:{x}=\mathrm{2arctan}\:\frac{{t}}{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{2}}}\:\rightarrow\:{dx}=\frac{\sqrt{\mathrm{2cos}^{\mathrm{3}} \:{x}}}{\mathrm{cos}\:\frac{{x}}{\mathrm{2}}}{dt}\right] \\ $$$$=\mathrm{2}\int\frac{{dt}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}}…
Question Number 211885 by Spillover last updated on 23/Sep/24 Answered by Frix last updated on 23/Sep/24 $$\int\frac{{x}^{\mathrm{2}} }{\left({x}\mathrm{sin}\:{x}\:+\mathrm{cos}\:{x}\right)^{\mathrm{2}} }{dx}=\frac{{y}}{{x}\mathrm{sin}\:{x}\:+\mathrm{cos}\:{x}}+{C} \\ $$$$\frac{{d}}{{dx}}\left[\frac{{y}}{{x}\mathrm{sin}\:{x}\:+\mathrm{cos}\:{x}}\right]=\frac{{x}^{\mathrm{2}} }{\left({x}\mathrm{sin}\:{x}\:+\mathrm{cos}\:{x}\right)^{\mathrm{2}} } \\ $$$$\frac{{y}'\left({x}\mathrm{sin}\:{x}\:+\mathrm{cos}\:{x}\right)−{yx}\mathrm{cos}\:{x}}{\left({x}\mathrm{sin}\:{x}\:+\mathrm{cos}\:{x}\right)^{\mathrm{2}}…
Question Number 211908 by Spillover last updated on 23/Sep/24 Answered by BHOOPENDRA last updated on 24/Sep/24 $${I}=\int_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} \left(\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{sin}\:\left(\frac{{k}\pi}{\mathrm{2}}\right){x}^{{k}} \right){dx} \\ $$$${The}\:{pattern}\:{for}\:\mathrm{sin}\:\left({k}\pi/\mathrm{2}\right) \\…
Question Number 211881 by Spillover last updated on 23/Sep/24 Answered by som(math1967) last updated on 23/Sep/24 $$\:\alpha+\beta+\gamma=\mathrm{1},\alpha\beta+\beta\gamma+\gamma\alpha=−\mathrm{2} \\ $$$$\:\alpha\beta\gamma=−\mathrm{1} \\ $$$$\:\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} +\gamma^{\mathrm{2}} =\mathrm{1}+\mathrm{4}=\mathrm{5} \\…
Question Number 211880 by Spillover last updated on 23/Sep/24 Answered by Frix last updated on 23/Sep/24 $$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{\mathrm{tan}\:{x}}{\pi^{\mathrm{2}} +\mathrm{4ln}^{\mathrm{2}} \:\mathrm{tan}\:{x}}{dx}\:\overset{\left[{t}=\mathrm{2ln}\:\mathrm{tan}\:{x}\right]} {=} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{−\infty} {\overset{\infty}…