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Category: Integration

Question-183031

Question Number 183031 by universe last updated on 18/Dec/22 Answered by aleks041103 last updated on 23/Dec/22 $$\left({x},{y}\right)\in{T}\left({G}\right) \\ $$$$\Rightarrow\frac{\pi}{\mathrm{2}}{s}\left(\mathrm{1}−{t}\right)={x} \\ $$$$\frac{\pi}{\mathrm{2}}\left(\mathrm{1}−{s}\right)={y}\Rightarrow{s}=\mathrm{1}−\frac{\mathrm{2}{y}}{\pi}\Rightarrow{x}=\left(\frac{\pi}{\mathrm{2}}−{y}\right)\left(\mathrm{1}−{t}\right) \\ $$$$\mathrm{0}<{s}<\mathrm{1}\Rightarrow\mathrm{0}<{y}<\pi/\mathrm{2} \\ $$$$\Rightarrow\mathrm{0}<{x}<\frac{\pi}{\mathrm{2}}−{y}…

Question-51910

Question Number 51910 by Meritguide1234 last updated on 01/Jan/19 Commented by Abdo msup. last updated on 01/Jan/19 $${let}\:{find}\:{firstI}_{{n}} =\:\int_{\mathrm{0}} ^{\pi} \:{cos}^{{n}} {x}\:{cos}\left({nx}\right){dx}\:{with}\:{n}\:{from}\:{N} \\ $$$${I}_{{n}} ={Re}\left(\:\int_{\mathrm{0}}…

Question-51876

Question Number 51876 by Tinkutara last updated on 31/Dec/18 Commented by maxmathsup by imad last updated on 31/Dec/18 $${let}\:{solve}\:{y}^{'} \:+\mathrm{2}{y}\:={f}\left({x}\right)\:\:\left({e}\right) \\ $$$$\left({he}\right)\:\Rightarrow{y}^{'} \:+\mathrm{2}{y}\:=\mathrm{0}\:\Rightarrow\frac{{y}^{'} }{{y}}=−\mathrm{2}\:\:\Rightarrow{ln}\left({y}\right)=−\mathrm{2}{x}\:+{k}\:\:\Rightarrow{y}\left({x}\right)={C}\:{e}^{−\mathrm{2}{x}} \\…