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By-using-the-substitution-x-cos-2-prove-that-1-x-1-x-dx-sin-2-2-C-

Question Number 152608 by ZiYangLee last updated on 30/Aug/21 $$\mathrm{By}\:\mathrm{using}\:\mathrm{the}\:\mathrm{substitution}\:{x}=\mathrm{cos}\:\mathrm{2}\theta, \\ $$$$\mathrm{prove}\:\mathrm{that}\:\int\:\sqrt{\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}}\:{dx}\:=\:−\mathrm{sin}\:\mathrm{2}\theta−\mathrm{2}\theta+{C} \\ $$ Answered by Olaf_Thorendsen last updated on 30/Aug/21 $$\mathrm{F}\left({x}\right)\:=\:\int\sqrt{\:\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}}\:{dx} \\ $$$$\mathrm{F}\left(\theta\right)\:=\:\int\sqrt{\:\frac{\mathrm{1}+\mathrm{cos2}\theta}{\mathrm{1}−\mathrm{cos2}\theta}}\:\left(−\mathrm{2sin2}\theta{d}\theta\right) \\…

Question-21503

Question Number 21503 by mondodotto@gmail.com last updated on 25/Sep/17 Answered by $@ty@m last updated on 26/Sep/17 $${Let}\:\mathrm{4}{x}^{\mathrm{2}} ={t} \\ $$$$\Rightarrow\mathrm{8}{xdx}={dt} \\ $$$${Change}\:{of}\:{limits}: \\ $$$${As}\:{x}\rightarrow\mathrm{4},\:{t}\rightarrow\mathrm{64} \\…

Question-21504

Question Number 21504 by mondodotto@gmail.com last updated on 25/Sep/17 Answered by $@ty@m last updated on 26/Sep/17 $${ATQ} \\ $$$$\boldsymbol{{S}}_{{n}} =\left(−\mathrm{1}\right)+\mathrm{1}+\mathrm{3}+\mathrm{5}+……+\left(\mathrm{2}{n}−\mathrm{3}\right) \\ $$$$\left({i}\right)\:\boldsymbol{{S}}_{\mathrm{50}} =\frac{\mathrm{50}}{\mathrm{2}}\left\{\left(−\mathrm{1}\right)+\left(\mathrm{2}×\mathrm{50}−\mathrm{3}\right)\right\} \\ $$$$=\mathrm{25}\left(−\mathrm{1}+\mathrm{97}\right)…

2log4-x-9logx-4-9-find-x-

Question Number 152571 by Gbenga last updated on 29/Aug/21 $$\:\mathrm{2}\boldsymbol{{log}}\mathrm{4}^{\boldsymbol{{x}}} +\mathrm{9}\boldsymbol{{logx}}^{\mathrm{4}} =\mathrm{9} \\ $$$$\boldsymbol{{find}}\:\boldsymbol{{x}} \\ $$ Answered by mr W last updated on 29/Aug/21 $$\mathrm{log}\:\left(\mathrm{4}^{\mathrm{2}{x}}…

k-1-n-k-5-

Question Number 152555 by Gbenga last updated on 29/Aug/21 $$\underset{\boldsymbol{{k}}=\mathrm{1}} {\overset{\boldsymbol{{n}}} {\boldsymbol{\sum}}{k}}^{\mathrm{5}} =? \\ $$ Answered by EDWIN88 last updated on 29/Aug/21 $$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left({k}+\mathrm{1}\right)^{\mathrm{5}}…

sin-x-sin-2x-sin-nx-cos-x-cos-2x-cos-nx-

Question Number 152551 by alisiao last updated on 29/Aug/21 $$\frac{{sin}\left({x}\right)+{sin}\left(\mathrm{2}{x}\right)+….+{sin}\left({nx}\right)}{{cos}\left({x}\right)+{cos}\left(\mathrm{2}{x}\right)+….+{cos}\left({nx}\right)}\:=\:? \\ $$ Answered by qaz last updated on 29/Aug/21 $$\frac{\mathrm{S}}{\mathrm{C}}=\frac{\Im\left(\mathrm{e}^{\mathrm{ix}} +\mathrm{e}^{\mathrm{2ix}} +…+\mathrm{e}^{\mathrm{nix}} \right)}{\Re\left(\mathrm{e}^{\mathrm{ix}} +\mathrm{e}^{\mathrm{2ix}} +…+\mathrm{e}^{\mathrm{nix}}…

Find-the-real-zeros-of-the-polynomial-P-a-x-x-2-1-x-1-2-ax-2-where-a-is-a-given-real-number-

Question Number 152544 by liberty last updated on 29/Aug/21 $$\:\mathrm{Find}\:\mathrm{the}\:\mathrm{real}\:\mathrm{zeros}\:\mathrm{of}\:\mathrm{the}\:\mathrm{polynomial} \\ $$$$\:\mathrm{P}_{\mathrm{a}} \left(\mathrm{x}\right)=\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{ax}^{\mathrm{2}} \\ $$$$\mathrm{where}\:\mathrm{a}\:\mathrm{is}\:\mathrm{a}\:\mathrm{given}\:\mathrm{real}\:\mathrm{number} \\ $$ Commented by MJS_new last updated on…

Question-21464

Question Number 21464 by mondodotto@gmail.com last updated on 24/Sep/17 Answered by myintkhaing last updated on 24/Sep/17 $$\mathrm{y}+\delta\mathrm{y}=\sqrt{\mathrm{tan}\left(\mathrm{x}+\delta\mathrm{x}\right)} \\ $$$$\delta\mathrm{y}=\sqrt{\mathrm{tan}\left(\mathrm{x}+\delta\mathrm{x}\right)}−\sqrt{\mathrm{tanx}} \\ $$$$\:\:\:\:\:=\left(\sqrt{\mathrm{tan}\left(\mathrm{x}+\delta\mathrm{x}\right)}−\sqrt{\mathrm{tanx}}\right)\frac{\sqrt{\mathrm{tan}\left(\mathrm{x}+\delta\mathrm{x}\right)}+\sqrt{\mathrm{tanx}}}{\:\sqrt{\mathrm{tan}\left(\mathrm{x}+\delta\mathrm{x}\right)}+\sqrt{\mathrm{tanx}}} \\ $$$$\:\:\:\:\:=\frac{\mathrm{tan}\left(\mathrm{x}+\delta\mathrm{x}\right)−\mathrm{tanx}}{\:\sqrt{\mathrm{tan}\left(\mathrm{x}+\delta\mathrm{x}\right)}+\sqrt{\mathrm{tanx}}}=\frac{\frac{\mathrm{tanx}+\mathrm{tan}\delta\mathrm{x}}{\mathrm{1}−\mathrm{tanxtan}\delta\mathrm{x}}−\mathrm{tanx}}{\mathrm{tan}\sqrt{\left(\mathrm{x}+\delta\mathrm{x}\right)}+\sqrt{\mathrm{tanx}}} \\ $$$$=\frac{\mathrm{tan}\delta\mathrm{x}\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}}…