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nice-calculus-prove-that-0-1-log-1-x-x-2-dx-2-2-

Question Number 137829 by mnjuly1970 last updated on 07/Apr/21 $$\:\:\:\:\:…….{nice}\:\:…\:…\:….\:{calculus}….. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{prove}\:{that}\::::: \\ $$$$\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\frac{{log}\left(\mathrm{1}−{x}\right)}{{x}}\right)^{\mathrm{2}} {dx}=\mathrm{2}\zeta\left(\mathrm{2}\right)…. \\ $$$$ \\ $$ Answered by EnterUsername last…

Question-72294

Question Number 72294 by mr W last updated on 27/Oct/19 Commented by mr W last updated on 27/Oct/19 $${the}\:{distances}\:{from}\:{a}\:{point}\:{to}\:{three} \\ $$$${vertices}\:{of}\:{a}\:{square}\:{are}\:{known}. \\ $$$${find}\:{the}\:{side}\:{length}\:{of}\:{the}\:{square}. \\ $$…

A-circle-of-radius-r-has-a-point-O-as-its-centre-Points-A-and-B-are-points-on-the-circumference-For-OAB-OA-OB-r-AB-d-AOB-What-is-r-d-

Question Number 6758 by FilupSmith last updated on 23/Jul/16 $$\mathrm{A}\:\mathrm{circle}\:\mathrm{of}\:\mathrm{radius}\:{r}\:\mathrm{has}\:\mathrm{a}\:\mathrm{point}\:{O}\:\mathrm{as}\:\mathrm{its} \\ $$$$\mathrm{centre}.\:\mathrm{Points}\:{A}\:\mathrm{and}\:{B}\:\mathrm{are}\:\mathrm{points}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{circumference}. \\ $$$$ \\ $$$$\mathrm{For}\:\bigtriangleup{OAB},\:\overline {{OA}}=\overline {{OB}}={r},\:\overline {{AB}}={d},\:\angle{AOB}=\theta. \\ $$$$\mathrm{What}\:\mathrm{is}\:\frac{{r}}{{d}}? \\ $$…

Question-72291

Question Number 72291 by aliesam last updated on 27/Oct/19 Answered by mr W last updated on 27/Oct/19 $$\left(\mathrm{5}{m}\:\mathrm{sin}\:\left(\mathrm{3}{x}\right)\right)^{\mathrm{2}} =\left(\mathrm{4}\:\mathrm{sin}\:{x}\right)^{\mathrm{2}} \\ $$$$\mathrm{5}{m}\:\mathrm{sin}\:\left(\mathrm{3}{x}\right)=\pm\mathrm{4}\:\mathrm{sin}\:{x} \\ $$$$\mathrm{5}{m}\:\mathrm{sin}\:{x}\:\left(\mathrm{3}−\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:{x}\right)=\pm\mathrm{4}\:\mathrm{sin}\:{x} \\…

lim-x-pi-2-1-tan-x-2-1-sin-x-1-tan-x-2-pi-2x-3-

Question Number 137820 by bramlexs22 last updated on 07/Apr/21 $$\underset{{x}\rightarrow\pi/\mathrm{2}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}−\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\right)\left(\mathrm{1}−\mathrm{sin}\:{x}\right)}{\left(\mathrm{1}+\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\right)\left(\pi−\mathrm{2}{x}\right)^{\mathrm{3}} }\:? \\ $$ Answered by SLVR last updated on 07/Apr/21 Commented by SLVR last…

Question-6750

Question Number 6750 by Tawakalitu. last updated on 22/Jul/16 Answered by Yozzii last updated on 22/Jul/16 $${Let}\:\angle{QPR}=\angle{RPS}=\alpha>\mathrm{0},\:\angle{PQR}=\beta>\mathrm{0}\:{and}\:\angle{PSR}=\delta>\mathrm{0}. \\ $$$${By}\:{the}\:{sine}\:{rule},\:{in}\:\bigtriangleup{QPR},\: \\ $$$$\frac{{QR}}{{sin}\alpha}=\frac{{QP}}{{sin}\angle{PRQ}}=\frac{\mathrm{3}{a}}{{sin}\left(\pi−\left(\alpha+\beta\right)\right)} \\ $$$$\Rightarrow{QR}=\frac{\mathrm{3}{asin}\alpha}{{sin}\left(\alpha+\beta\right)}. \\ $$$${In}\:\bigtriangleup{PRS},\:{by}\:{the}\:{sine}\:{rule}\:{we}\:{have}…