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Author: Tinku Tara

Question-60745

Question Number 60745 by Tawa1 last updated on 25/May/19 Commented by Prithwish sen last updated on 25/May/19 $$\frac{\left[\left(\mathrm{n}+\mathrm{1}\right)!\right]^{\mathrm{n}} }{\left(\mathrm{n}!\right)^{\mathrm{n}+\mathrm{1}} }\:=\left[\frac{\left(\mathrm{n}+\mathrm{1}\right)!}{\mathrm{n}!}\right]^{\mathrm{n}} .\frac{\mathrm{1}}{\mathrm{n}!}\:=\frac{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}!} \\ $$$$=\frac{\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{1}}.\frac{\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{2}}…………\frac{\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{n}}\:>\mathrm{1} \\…

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…