Question Number 152266 by mathdanisur last updated on 26/Aug/21 Answered by qaz last updated on 27/Aug/21 $$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{Li}_{\mathrm{2}} \left(\mathrm{x}\right)\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)\mathrm{dx} \\ $$$$=\left[\left(\mathrm{1}+\mathrm{x}\right)\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)−\left(\mathrm{1}+\mathrm{x}\right)\right]\mathrm{Li}_{\mathrm{2}} \left(\mathrm{x}\right)\mid_{\mathrm{0}} ^{\mathrm{1}} +\int_{\mathrm{0}}…
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Question Number 86723 by A8;15: last updated on 30/Mar/20 Commented by jagoll last updated on 30/Mar/20 $$\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}−\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}\:=\:\mathrm{tan}\:\mathrm{x}\:×\:\mathrm{cot}\:\mathrm{x} \\ $$$$\mathrm{cos}\:\mathrm{2x}\:=\:\mathrm{1}\:\Rightarrow\:\begin{cases}{\mathrm{x}\:=\:\mathrm{0}}\\{\mathrm{x}\:=\:\pi}\end{cases} \\ $$ Terms of…
Question Number 86716 by john santu last updated on 30/Mar/20 $$\mathrm{cos}\:\left(\frac{\pi}{\mathrm{17}}\right)×\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{17}}\right)×\mathrm{cos}\:\left(\frac{\mathrm{4}\pi}{\mathrm{17}}\right)×\mathrm{cos}\:\left(\frac{\mathrm{8}\pi}{\mathrm{17}}\right)\:= \\ $$ Commented by jagoll last updated on 30/Mar/20 $$\frac{\mathrm{cos}\:\mathrm{x}\:\mathrm{cos}\:\mathrm{2x}\:\mathrm{cos}\:\mathrm{4x}\:\mathrm{cos}\:\mathrm{8x}}{\mathrm{2sin}\:\mathrm{x}}\:×\:\mathrm{2sin}\:\mathrm{x}\:\: \\ $$$$\frac{\mathrm{sin}\:\mathrm{2x}\:\mathrm{cos}\:\mathrm{2x}\:\mathrm{cos}\:\mathrm{4x}\:\mathrm{cos}\:\mathrm{8x}}{\mathrm{2sin}\:\mathrm{x}}\:=\: \\ $$$$\frac{\mathrm{sin}\:\mathrm{4x}\:\mathrm{cos}\:\mathrm{4x}\:\mathrm{cos}\:\mathrm{8x}}{\mathrm{4sin}\:\mathrm{x}}\:=\:…
Question Number 21179 by Joel577 last updated on 15/Sep/17 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{\mathrm{16}{x}^{\mathrm{2}} \:+\:\mathrm{4}{x}}\:−\:\sqrt{{x}^{\mathrm{2}} }\:−\:\sqrt{\mathrm{9}{x}^{\mathrm{2}} \:+\:\mathrm{3}{x}} \\ $$ Answered by dioph last updated on 15/Sep/17 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}\sqrt{\mathrm{16}+\frac{\mathrm{4}}{{x}}}\:−\:{x}\:−\:{x}\sqrt{\mathrm{9}+\frac{\mathrm{3}}{{x}}}…
Question Number 152244 by mnjuly1970 last updated on 26/Aug/21 $$ \\ $$$$\:\:\:\:\:\:\boldsymbol{\phi}\::=\:\int_{\mathrm{0}} ^{\:\infty} \frac{\:{cos}\:\left({x}\:\right).{cosh}\:\left({x}\:\right)}{{cosh}\:\left(\pi{x}\:\right)}{dx}=? \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 152247 by mathocean1 last updated on 26/Aug/21 $${show}\:{that}\:\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\frac{\frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}}\:−\mathrm{1}}{{x}}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$ Answered by mr W last updated on 26/Aug/21 $$\mathrm{ln}\:\left(\mathrm{1}+{x}\right)={x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−\frac{{x}^{\mathrm{4}} }{\mathrm{4}}+……
Question Number 86708 by jagoll last updated on 31/Mar/20 $$\int\mathrm{x}\:\sqrt{\sqrt{\mathrm{2}}\:\mathrm{x}−\sqrt{\mathrm{2x}^{\mathrm{2}} −\mathrm{1}}}\:\mathrm{dx}\: \\ $$ Answered by john santu last updated on 31/Mar/20 $$\mathrm{let}\::\:\sqrt{\mathrm{2}\:}\mathrm{x}\:−\sqrt{\mathrm{2x}^{\mathrm{2}} −\mathrm{1}}\:=\:\mathrm{u}\:\left(\mathrm{i}\right) \\ $$$$\Rightarrow\:\frac{\mathrm{2x}^{\mathrm{2}}…
Question Number 152241 by ZiYangLee last updated on 26/Aug/21 Answered by Olaf_Thorendsen last updated on 26/Aug/21 $$\left({i}\right)\:\mathrm{and}\:\left({ii}\right)\:: \\ $$$${p}\left({x}\right)\:=\:{a}\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +{px}+{q}\right),\:{a}\neq\mathrm{0} \\ $$$$\left({iii}\right)\:: \\ $$$${p}\left(\mathrm{0}\right)\:=\:\mathrm{4}\:\Leftrightarrow\:−{aq}\:=\:\mathrm{4}\:\Rightarrow\:{q}\:=\:−\frac{\mathrm{4}}{{a}} \\…
Question Number 152240 by mnjuly1970 last updated on 26/Aug/21 $$ \\ $$$$\:\:{prove}\:{that}.. \\ $$$$\:\:\:{csch}\:\left({x}\right)=\:\frac{\mathrm{1}}{{x}}\:+\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}.\left(−\mathrm{1}\right)^{\:{n}} \:{x}}{{n}^{\:\mathrm{2}} \pi^{\:\mathrm{2}} +\:{x}^{\:\mathrm{2}} } \\ $$$$\:\:\:{then}\:{find}: \\ $$$$\:\:\:\:\Omega\::=\:\int_{\mathrm{0}} ^{\:\infty}…