Question Number 69735 by mhmd last updated on 27/Sep/19 $$\underset{\:\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\:\frac{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}{\mathrm{3}+\mathrm{sin}\:\mathrm{2}{x}}\:{dx}\:= \\ $$ Commented by mathmax by abdo last updated on 27/Sep/19 $${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}}…
Question Number 69328 by mhmd last updated on 22/Sep/19 $$\underset{\:\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\:\frac{\phi\left({x}\right)}{\phi\left({x}\right)+\phi\left(\frac{\pi}{\mathrm{2}}\:−{x}\right)}\:{dx}\:= \\ $$ Commented by mathmax by abdo last updated on 22/Sep/19 $${let}\:{I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}}…
Question Number 69327 by mhmd last updated on 22/Sep/19 $$\:\underset{\:\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\:\frac{\mathrm{2}^{\mathrm{2}{x}+\mathrm{1}} −\:\mathrm{5}^{\mathrm{2}{x}−\mathrm{1}} }{\mathrm{10}^{{x}} }\:{dx}\:=\: \\ $$ Answered by MJS last updated on 22/Sep/19 $$\int\frac{\mathrm{2}^{\mathrm{2}{x}+\mathrm{1}}…
Question Number 68352 by mhmd last updated on 09/Sep/19 $$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\:\frac{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \mathrm{15}°}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{15}°}\:\:\mathrm{is} \\ $$ Answered by $@ty@m123 last updated on 09/Sep/19 $$\frac{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \theta}{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \theta}=\mathrm{cos}\:\mathrm{2}\theta \\…
Question Number 68353 by mhmd last updated on 09/Sep/19 $$\mathrm{If}\:\mathrm{sin}\:{x}+\mathrm{sin}^{\mathrm{2}} {x}=\mathrm{1},\:\mathrm{then}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{cos}^{\mathrm{2}} {x}+\mathrm{cos}^{\mathrm{4}} {x}\:\:\mathrm{is} \\ $$ Answered by $@ty@m123 last updated on 09/Sep/19 $${Given}\:,\:\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}}…
Question Number 68351 by mhmd last updated on 09/Sep/19 $$\mathrm{If}\:\mathrm{2}{x}^{\mathrm{2}} +\left(\mathrm{2}{p}−\mathrm{13}\right){x}+\mathrm{2}=\mathrm{0}\:\mathrm{is}\:\mathrm{exactly}\:\mathrm{divisible} \\ $$$$\mathrm{by}\:{x}−\mathrm{3},\:\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{p}\:\mathrm{is} \\ $$ Answered by $@ty@m123 last updated on 09/Sep/19 $${f}\left({x}\right)=\mathrm{2}{x}^{\mathrm{2}} +\left(\mathrm{2}{p}−\mathrm{13}\right){x}+\mathrm{2} \\…
Question Number 68151 by mhmd last updated on 06/Sep/19 $$\mathrm{If}\:\mid{a}\mid<\:\mathrm{1}\:\mathrm{and}\:\mid{b}\mid<\:\mathrm{1},\:\mathrm{then}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{series} \\ $$$$\mathrm{1}+\left(\mathrm{1}+{a}\right){b}+\left(\mathrm{1}+{a}+{a}^{\mathrm{2}} \right){b}^{\mathrm{2}} +\left(\mathrm{1}+{a}+{a}^{\mathrm{2}} +{a}^{\mathrm{3}} \right){b}^{\mathrm{3}} +… \\ $$$$\mathrm{is} \\ $$ Commented by ~ À…
Question Number 67836 by gunawan last updated on 01/Sep/19 $$\mathrm{If}\:\begin{vmatrix}{{x}^{{n}} }&{{x}^{{n}+\mathrm{2}} }&{{x}^{{n}+\mathrm{3}} }\\{{y}^{{n}} }&{{y}^{{n}+\mathrm{2}} }&{{y}^{{n}+\mathrm{3}} }\\{{z}^{{n}} }&{{z}^{{n}+\mathrm{2}} }&{{z}^{{n}+\mathrm{3}} }\end{vmatrix} \\ $$$$=\:\left({x}−{y}\right)\left({y}−{z}\right)\left({z}−{x}\right)\left(\frac{\mathrm{1}}{{x}}\:+\:\frac{\mathrm{1}}{{y}}\:+\:\frac{\mathrm{1}}{{z}}\right), \\ $$$$\mathrm{then}\:{n}\:\mathrm{equals} \\ $$…
Question Number 66090 by Lalita 42@gmeilcon last updated on 09/Aug/19 $$\:\underset{\:\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\:\frac{\mathrm{sin}^{\mathrm{2}} {x}}{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}\:{dx}\:= \\ $$ Answered by mr W last updated on 09/Aug/19 $$\:{I}=\underset{\:\mathrm{0}}…
Question Number 193391 by DAVONG last updated on 12/Jun/23 Answered by aba last updated on 12/Jun/23 $$\left(\mathrm{1}\right)\:\mathrm{log}_{\mathrm{a}} \left(\mathrm{6}\right)−\mathrm{log}_{\mathrm{a}} \left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:\mathrm{log}_{\mathrm{a}} \left(\frac{\mathrm{6}}{\mathrm{x}}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{ln}\left(\frac{\mathrm{6}}{\mathrm{x}}\right)=\mathrm{ln}\left(\sqrt{\mathrm{a}}\right)\:\Rightarrow\:\mathrm{x}=\frac{\mathrm{6}}{\:\sqrt{\mathrm{a}}}\:\checkmark \\ $$ Commented…