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Category: Integration

Question-126274

Question Number 126274 by bramlexs22 last updated on 19/Dec/20 Answered by liberty last updated on 19/Dec/20 $$\:{letting}\:{x}=\sqrt{\mathrm{2}}\:\mathrm{sec}\:\ell\:{with}\:\rightarrow\begin{cases}{\ell=\frac{\pi}{\mathrm{2}}}\\{\ell=\frac{\pi}{\mathrm{4}}}\end{cases} \\ $$$$\:\int_{\pi/\mathrm{4}} ^{\:\pi/\mathrm{2}} \:\frac{\sqrt{\mathrm{2}}\:\mathrm{sec}\:\ell\:\mathrm{tan}\:\ell}{\:\sqrt{\mathrm{2}}\:\mathrm{sec}\:\ell\:\sqrt{\mathrm{2tan}\:^{\mathrm{2}} \ell}}\:{d}\ell\:= \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\:{d}\ell\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\left[\:\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\:\right]\:=\:\frac{\pi}{\mathrm{4}\sqrt{\mathrm{2}}}\:=\:\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{8}}\: \\…

evaluate-i-x-1-x-1-dx-ii-0-pi-2cosxsinx-dx-iii-pi-3-pi-sin2x-cos2x-dx-

Question Number 60739 by Forkum Michael Choungong last updated on 25/May/19 $${evaluate}\:\: \\ $$$${i}.\int\:\left(\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}\right){dx} \\ $$$${ii}.\:\:\int_{\mathrm{0}} ^{\pi} \left(\mathrm{2}{cosxsinx}\right){dx}\:\: \\ $$$${iii}.\:\:\int_{\frac{\pi}{\mathrm{3}\:}\:} ^{\pi} \left(\frac{{sin}\mathrm{2}{x}}{{cos}\mathrm{2}{x}}\right){dx} \\ $$ Commented…

Question-60716

Question Number 60716 by peter frank last updated on 24/May/19 Commented by maxmathsup by imad last updated on 25/May/19 $${let}\:{A}\left({x}\right)\:=\left(\frac{{x}}{{x}+\mathrm{1}}\right)^{{x}} \:\Rightarrow{A}\left({x}\right)\:=\left(\mathrm{1}−\frac{{x}+\mathrm{1}}{{x}}\right)^{{x}} \:={e}^{{x}\:{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}+\mathrm{1}}\right)} \\ $$$${but}\:{we}\:{have}\:{for}\:{u}\:\in{V}\left(\mathrm{0}\right)\:{ln}\left(\mathrm{1}−{u}\right)\:\sim−{u}\:\Rightarrow{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}+\mathrm{1}}\right)\sim−\frac{\mathrm{1}}{{x}+\mathrm{1}}\:{for}\:{x}\in{V}\left(+\infty\right) \\…