Question Number 109016 by mathdave last updated on 20/Aug/20 Answered by Dwaipayan Shikari last updated on 20/Aug/20 $$\int\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{sinx}}}{dx} \\ $$$$\int\frac{\mathrm{1}}{{cos}\frac{{x}}{\mathrm{2}}−{sin}\frac{{x}}{\mathrm{2}}}{dx} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{\mathrm{1}}{{cos}\left(\frac{{x}}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)}{dx} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int{sec}\left(\frac{{x}}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right){dx} \\…
Question Number 109017 by mathdave last updated on 20/Aug/20 Answered by mathmax by abdo last updated on 20/Aug/20 $$\mathrm{I}\:=\int_{\mathrm{arctan}\left(\mathrm{1}\right)} ^{\mathrm{arctan}\left(\sqrt{\mathrm{2}}\right)} \sqrt{\frac{\mathrm{8}}{\mathrm{1}−\mathrm{sin}\left(\mathrm{4x}\right)}}\mathrm{dx}\:\Rightarrow\mathrm{I}\:=\int_{\mathrm{arctan}\left(\mathrm{1}\right)} ^{\mathrm{arctan}\left(\sqrt{\mathrm{2}}\right)} \sqrt{\frac{\mathrm{8}}{\mathrm{1}−\mathrm{cos}\left(\frac{\pi}{\mathrm{2}}−\mathrm{4x}\right)}}\mathrm{dx} \\ $$$$=\int_{\mathrm{arctan}\left(\mathrm{1}\right)}…
Question Number 43452 by mondodotto@gmail.com last updated on 10/Sep/18 $$\mathrm{prove}\:\mathrm{that} \\ $$$$\boldsymbol{\mathrm{tanh}{x}}=\boldsymbol{{i}\mathrm{tan}{x}} \\ $$ Answered by alex041103 last updated on 10/Sep/18 $$\mathrm{tanh}\:{x}\:=\:\frac{{e}^{{x}} −{e}^{−{x}} }{{e}^{{x}} +{e}^{−{x}}…
Question Number 43454 by mondodotto@gmail.com last updated on 10/Sep/18 $$\mathrm{qn}:\:\mathrm{There}\:\mathrm{is}\:\mathrm{a}\:\mathrm{group}\:\mathrm{of}\:\mathrm{50}\:\mathrm{people} \\ $$$$\mathrm{who}\:\mathrm{are}\:\mathrm{patriotic}\:\mathrm{out}\:\mathrm{of}\:\mathrm{which}\:\mathrm{20} \\ $$$$\mathrm{believes}\:\mathrm{in}\:\mathrm{non}\:\mathrm{violence}.\:\mathrm{Two}\:\mathrm{persons} \\ $$$$\mathrm{are}\:\mathrm{selected}\:\mathrm{at}\:\mathrm{rondom}\:\mathrm{out}\:\mathrm{of}\:\mathrm{them}. \\ $$$$\mathrm{write}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{distribution}\:\mathrm{for}\:\mathrm{the} \\ $$$$\mathrm{selected}\:\mathrm{persons}\:\mathrm{who}\:\mathrm{are}\:\mathrm{non}\:\mathrm{violent}. \\ $$$$\mathrm{also}\:\mathrm{find}\:\mathrm{the}\:\mathrm{mean}\:\mathrm{of}\:\mathrm{the}\:\mathrm{distribution}. \\ $$$$\mathrm{explain}\:\mathrm{the}\:\mathrm{importance}\:\mathrm{of}\:\mathrm{non}\:\mathrm{violence}\:\mathrm{in}\:\mathrm{patriotism}. \\…
Question Number 174519 by Ml last updated on 03/Aug/22 Commented by Michaelfaraday last updated on 17/Aug/22 $${please}\:{can}\:{you}\:{show}\:{the}\:{so}<{ution}\:{of} \\ $$$${your}\:{answer}. \\ $$ Commented by Rasheed.Sindhi last…
Question Number 174508 by Gbenga last updated on 03/Aug/22 $$ \\ $$$$\mathscr{L}\left(\boldsymbol{\mathrm{sin}}^{\boldsymbol{\mathrm{n}}} \left(\boldsymbol{\mathrm{x}}\right)\right)=? \\ $$ Answered by Mathspace last updated on 03/Aug/22 $${sin}^{{n}} {x}=\left(\frac{{e}^{{ix}} −{e}^{−{ix}}…
Question Number 174514 by Best1 last updated on 03/Aug/22 $${a}+{b}+{c}=\mathrm{1} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{2} \\ $$$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} =\mathrm{3} \\ $$$${then}\:{a}^{\mathrm{5}} +{b}^{\mathrm{5}} +{c}^{\mathrm{5}} \:?…
Question Number 174492 by naka3546 last updated on 02/Aug/22 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{\mathrm{3}{x}\:\mathrm{tan}\:\frac{\mathrm{2}}{{x}}\:−\:\mathrm{2}{x}\:\mathrm{sin}\:\frac{\mathrm{3}}{{x}}}{\mathrm{cos}\:\frac{\mathrm{1}}{{x}}\:−\:\mathrm{cos}\:\frac{\mathrm{2}}{{x}}}\:=\:\:? \\ $$ Commented by kaivan.ahmadi last updated on 02/Aug/22 $${we}\:{use}\:{Taylor}\:{series} \\ $$$$\sim{li}\underset{{x}\rightarrow\infty} {{m}}\:\frac{\mathrm{3}{x}\left(\frac{\mathrm{2}}{{x}}−\frac{\mathrm{8}}{\mathrm{3}{x}^{\mathrm{3}} }\right)−\mathrm{2}{x}\left(\frac{\mathrm{3}}{{x}}−\frac{\mathrm{9}}{\mathrm{2}{x}^{\mathrm{3}}…
Question Number 108961 by ZiYangLee last updated on 20/Aug/20 $$\mathrm{Solve}\:\frac{\mid{x}−\mathrm{2}\mid+\mathrm{1}}{\mid{x}−\mathrm{2}\mid−\mathrm{1}}<\mathrm{3} \\ $$ Commented by Rasheed.Sindhi last updated on 20/Aug/20 $$\:\frac{\mid{x}−\mathrm{2}\mid+\mathrm{1}}{\mid{x}−\mathrm{2}\mid−\mathrm{1}}<\mathrm{3} \\ $$$$\:\frac{\mid{x}−\mathrm{2}\mid+\mathrm{1}}{\mid{x}−\mathrm{2}\mid−\mathrm{1}}<\frac{\mathrm{2}+\mathrm{1}}{\mathrm{2}−\mathrm{1}} \\ $$$$\:\frac{\mid{x}−\mathrm{2}\mid}{\mid{x}−\mathrm{2}\mid}<\frac{\mathrm{2}}{\mathrm{2}} \\…
Question Number 108950 by ZiYangLee last updated on 20/Aug/20 $$\mathrm{In}\:\bigtriangleup\mathrm{ABC},\: \\ $$$$\mathrm{prove}\:\mathrm{that}\:\frac{\mathrm{1}+\mathrm{cosA}−\mathrm{cosB}+\mathrm{cosC}}{\mathrm{1}+\mathrm{cosA}+\mathrm{cosB}−\mathrm{cosC}}=\mathrm{tan}\frac{\mathrm{B}}{\mathrm{2}}\mathrm{cot}\frac{\mathrm{C}}{\mathrm{2}} \\ $$ Answered by mnjuly1970 last updated on 20/Aug/20 $$\frac{\left(\mathrm{1}+{cos}\left(\mathrm{C}\right)\right)+\left({cos}\left(\mathrm{A}\right)−{cos}\left(\mathrm{B}\right)\right)}{\left(\mathrm{1}−{cos}\left(\mathrm{C}\right)\right)+\left({cos}\left(\mathrm{A}\right)+{cos}\left(\mathrm{B}\right)\right)} \\ $$$$=\frac{\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{\mathrm{C}}{\mathrm{2}}\right)−\mathrm{2}{sin}\left(\frac{\mathrm{A}+\mathrm{B}}{\mathrm{2}}\right){sin}\left(\frac{\mathrm{A}−\mathrm{B}}{\mathrm{2}}\right)}{\mathrm{2}{sin}^{\mathrm{2}}…