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Question-197238

Question Number 197238 by sonukgindia last updated on 10/Sep/23 Commented by Frix last updated on 12/Sep/23 $$\mathrm{No}\:\mathrm{useful}\:\mathrm{exact}\:\mathrm{solution}. \\ $$$$\mathrm{Set}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{quarter}\:\mathrm{circle}\:=\mathrm{1} \\ $$$$\Rightarrow \\ $$$$\mathrm{Radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{semi}\:\mathrm{circle}\:\approx.\mathrm{717910784} \\ $$$${r}\approx.\mathrm{205269551}…

Question-197225

Question Number 197225 by sonukgindia last updated on 10/Sep/23 Answered by Frix last updated on 10/Sep/23 $${x}=\sqrt[{\mathrm{2}}]{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{10}}{\mathrm{9}}{x}}} \\ $$$${x}^{\mathrm{2}} =\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{10}}{\mathrm{9}}{x}} \\ $$$$\mathrm{3}\left({x}^{\mathrm{2}} −\mathrm{1}\right)=\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{10}}{\mathrm{9}}{x}} \\ $$$$\mathrm{27}\left({x}^{\mathrm{2}}…

Question-197229

Question Number 197229 by sonukgindia last updated on 10/Sep/23 Commented by Sachinkhar last updated on 10/Sep/23 $$\boldsymbol{\mathrm{Solve}}\:\boldsymbol{\mathrm{fourier}}\:\boldsymbol{\mathrm{inverse}}\:\boldsymbol{\mathrm{transform}}\:\boldsymbol{\mathrm{of}}\: \\ $$$$\boldsymbol{\mathrm{F}}\left(\boldsymbol{\xi}\right)=\boldsymbol{\mathrm{cosat}\xi}^{\mathrm{2}} \\ $$ Answered by HeferH last…

Question-197226

Question Number 197226 by sonukgindia last updated on 10/Sep/23 Answered by Frix last updated on 10/Sep/23 $$\int\frac{{dx}}{\mathrm{1}+\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}\:\overset{{t}=\frac{\mathrm{2}{x}+\mathrm{1}+\mathrm{2}\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}{\:\sqrt{\mathrm{3}}}} {=} \\ $$$$=\int\frac{{t}^{\mathrm{2}} +\mathrm{1}}{{t}\left({t}^{\mathrm{2}} +\frac{\mathrm{4}{t}}{\:\sqrt{\mathrm{3}}}+\mathrm{1}\right)}{dt}=\int\left(\frac{\mathrm{1}}{{t}}+\frac{\mathrm{2}}{{t}+\sqrt{\mathrm{3}}}−\frac{\mathrm{6}}{\mathrm{3}{t}+\sqrt{\mathrm{3}}}\right){dt}= \\…