Question Number 149805 by mnjuly1970 last updated on 07/Aug/21 Answered by Ar Brandon last updated on 07/Aug/21 $${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{csc}^{\mathrm{2}} {x}\mathrm{ln}\left(\mathrm{1}+\mathrm{2sin}^{\mathrm{2}} {x}\right){dx} \\ $$$$\begin{cases}{{u}\left({x}\right)=\mathrm{ln}\left(\mathrm{1}+\mathrm{2sin}^{\mathrm{2}} {x}\right)}\\{{v}'\left({x}\right)=\mathrm{csc}^{\mathrm{2}}…
Question Number 149796 by mr W last updated on 07/Aug/21 Commented by mr W last updated on 07/Aug/21 $${solution}\:{to}\:{Q}\mathrm{148806} \\ $$ Commented by mr W…
Question Number 84263 by Rio Michael last updated on 10/Mar/20 $$\mathrm{Using}\:\mathrm{the}\:\mathrm{approximation} \\ $$$$\:{h}\left(\frac{{dy}}{{dx}}\right)_{{n}} \:\approx\:{y}_{{n}+\mathrm{1}} −{y}_{{n}} \:\mathrm{and}\:\mathrm{that}\:\frac{{dy}}{{dx}}\:=\:\mathrm{1},\:{y}\:=\mathrm{2} \\ $$$$\mathrm{when}\:{x}\:=\:\mathrm{0}\:.\:\mathrm{then}\:,\:{y}_{\mathrm{1}} \:= \\ $$$$\left[\mathrm{A}\right]\:{h}−\mathrm{2} \\ $$$$\left[\mathrm{B}\right]\:{h}\:+\:\mathrm{2} \\ $$$$\left[\mathrm{C}\right]\:{h}−\mathrm{1}…
Question Number 84261 by 698148290 last updated on 10/Mar/20 Answered by Rio Michael last updated on 10/Mar/20 $$\:^{{n}} {C}_{{r}} =\:\frac{{n}!}{\left({n}−{r}\right)!{r}!} \\ $$$$\:^{{n}} {C}_{\mathrm{12}} \:=\:^{{n}} {C}_{\mathrm{8}}…
Question Number 149793 by mathdanisur last updated on 07/Aug/21 $$\mathrm{Compare}: \\ $$$$\mathrm{tan}\left(\mathrm{11}°\right)\:\:\:\mathrm{and}\:\:\:\frac{\mathrm{1}}{\mathrm{5}} \\ $$ Commented by DonQuichote last updated on 07/Aug/21 $$\frac{\mathrm{1}}{\mathrm{5}}\in\mathbb{Q}\:{but}\:{tan}\:\mathrm{11}°\:\notin\mathbb{Q} \\ $$$${what}\:{else}\:{can}\:{I}\:{tell}\:{you}? \\…
Question Number 18723 by mondodotto@gmail.com last updated on 28/Jul/17 Answered by Tinkutara last updated on 29/Jul/17 $$\mathrm{2sin2}\theta\:+\:\mathrm{15cos2}\theta\:=\:\mathrm{10} \\ $$$$\frac{\mathrm{2sin2}\theta}{\:\sqrt{\mathrm{229}}}\:+\:\frac{\mathrm{15cos2}\theta}{\:\sqrt{\mathrm{229}}}\:=\:\frac{\mathrm{10}}{\:\sqrt{\mathrm{229}}} \\ $$$$\mathrm{cos}\left(\mathrm{2}\theta\:−\:\alpha\right)\:=\:\frac{\mathrm{10}}{\:\sqrt{\mathrm{229}}}\:,\:\mathrm{where}\:\alpha\:=\:\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{15}}{\:\sqrt{\mathrm{229}}}\right) \\ $$$$\theta\:=\:{n}\pi\:\pm\:\frac{\mathrm{5}}{\:\sqrt{\mathrm{229}}}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{15}}{\:\sqrt{\mathrm{229}}}\right)…
Question Number 84259 by M±th+et£s last updated on 10/Mar/20 Commented by M±th+et£s last updated on 10/Mar/20 $${find}\:\frac{{y}}{{x}} \\ $$$${and}\:{z} \\ $$ Answered by mr W…
Question Number 18722 by mondodotto@gmail.com last updated on 28/Jul/17 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 149795 by mathdanisur last updated on 07/Aug/21 $$\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}\centerdot\mathrm{3}\centerdot\mathrm{5}\centerdot\mathrm{7}\centerdot\:…\:\centerdot\left(\mathrm{2n}-\mathrm{1}\right)}{\mathrm{2}\centerdot\mathrm{4}\centerdot\mathrm{6}\centerdot\:…\:\centerdot\mathrm{2n}}\:=\:? \\ $$ Answered by mathmax by abdo last updated on 07/Aug/21 $$\mathrm{u}_{\mathrm{n}} =\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}.\mathrm{7}…..\left(\mathrm{2n}−\mathrm{1}\right)}{\mathrm{2}.\mathrm{4}.\mathrm{6}….\left(\mathrm{2n}\right)}\:=\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}….\left(\mathrm{2n}−\mathrm{1}\right)}{\mathrm{2}^{\mathrm{n}} \:\mathrm{n}!}…
Question Number 18721 by Tinkutara last updated on 28/Jul/17 $$\mathrm{Let}\:{ABC}\:\mathrm{and}\:{ABC}'\:\mathrm{be}\:\mathrm{two}\:\mathrm{non}- \\ $$$$\mathrm{congruent}\:\mathrm{triangles}\:\mathrm{with}\:\mathrm{sides}\:{AB}\:=\:\mathrm{4}, \\ $$$${AC}\:=\:{AC}'\:=\:\mathrm{2}\sqrt{\mathrm{2}}\:\mathrm{and}\:\mathrm{angle}\:{B}\:=\:\mathrm{30}°. \\ $$$$\mathrm{The}\:\mathrm{absolute}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{difference} \\ $$$$\mathrm{between}\:\mathrm{the}\:\mathrm{areas}\:\mathrm{of}\:\mathrm{these}\:\mathrm{triangles}\:\mathrm{is} \\ $$ Answered by ajfour last updated…