Menu Close

Category: None

1-i-4i-

Question Number 48227 by gunawan last updated on 21/Nov/18 $$\left(\mathrm{1}−{i}\right)^{\mathrm{4}{i}} =.. \\ $$ Answered by Smail last updated on 21/Nov/18 $$\left(\mathrm{1}−{i}\right)^{\mathrm{4}{i}} =\left(\sqrt{\mathrm{2}}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}−{i}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)\right)^{\mathrm{4}{i}} =\left(\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} \right)^{\mathrm{4}{i}} \left({e}^{−\frac{{i}\pi}{\mathrm{4}}}…

calculate-log-1-3-i-2-

Question Number 48226 by gunawan last updated on 21/Nov/18 $$\mathrm{calculate} \\ $$$$\mathrm{log}\left(−\mathrm{1}+\sqrt{\mathrm{3}}\:\mathrm{i}\right)^{\mathrm{2}} \\ $$ Answered by Smail last updated on 21/Nov/18 $${ln}\left(\left(−\mathrm{1}+\sqrt{\mathrm{3}}{i}\right)^{\mathrm{2}} \right)=\mathrm{2}{ln}\left(\mathrm{2}\left(−\frac{\mathrm{1}}{\mathrm{2}}+{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\right) \\ $$$$=\mathrm{2}{ln}\mathrm{2}+\mathrm{2}{ln}\left({cos}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)+{isin}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\right)…

f-x-x-1-3-2f-x-1-f-10-

Question Number 48208 by naka3546 last updated on 20/Nov/18 $${f}\left({x}\right)\:+\:\left({x}\:+\:\mathrm{1}\right)^{\mathrm{3}} \:\:=\:\:\mathrm{2}{f}\left({x}\:+\:\mathrm{1}\right) \\ $$$${f}\left(\mathrm{10}\right)\:\:=\:\:? \\ $$ Answered by MJS last updated on 20/Nov/18 $${f}\left({x}\right)={ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d}…

Question-113714

Question Number 113714 by mathdave last updated on 14/Sep/20 Answered by Olaf last updated on 15/Sep/20 $$\mathrm{I}_{{n}} \:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{n}} \mathrm{ln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}}{dx} \\ $$$$\mathrm{I}_{\mathrm{0}} \:=\:\int_{\mathrm{0}} ^{\mathrm{1}}…

tanx-dx-

Question Number 113706 by Rasikh last updated on 15/Sep/20 $$\:\:\:\int\sqrt{\mathrm{tanx}}\:\mathrm{dx}\:=?\:\:\:\: \\ $$ Answered by MJS_new last updated on 15/Sep/20 $$\int\sqrt{\mathrm{tan}\:{x}}\:{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{\mathrm{tan}\:{x}}\:\rightarrow\:{dx}=\mathrm{2cos}^{\mathrm{2}} \:{x}\:\sqrt{\mathrm{tan}\:{x}}\:{dt}\right] \\ $$$$=\mathrm{2}\int\frac{{t}^{\mathrm{2}}…