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Category: Integration

cot-1-x-dx-

Question Number 94119 by i jagooll last updated on 17/May/20 $$\int\:\mathrm{cot}^{−\mathrm{1}} \left(\sqrt{\mathrm{x}}\right)\:\mathrm{dx}\: \\ $$ Commented by i jagooll last updated on 17/May/20 $${u}\:=\:\mathrm{cot}^{−\mathrm{1}} \left(\sqrt{{x}}\right)\:\Rightarrow\:{du}\:=\:\frac{{dx}}{\mathrm{2}\sqrt{{x}}\:\left(\mathrm{1}−{x}\right)} \\…

evaluate-the-inequality-for-n-2-pi-2-1-n-1-n-1-n-lt-1-n-pi-2-sin-t-1-n-dt-

Question Number 94093 by MAB last updated on 16/May/20 $${evaluate}\:{the}\:{inequality}\:{for}\:{n}\geqslant\mathrm{2} \\ $$$$\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{1}}{{n}}\right)\frac{\mathrm{1}}{\:\sqrt[{{n}}]{{n}}}<\int_{\frac{\mathrm{1}}{{n}}} ^{\frac{\pi}{\mathrm{2}}} \sqrt[{{n}}]{{sin}\left({t}\right)}{dt} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

cotx-dx-

Question Number 94084 by seedhamaieng@gmail.com last updated on 16/May/20 $$\int\sqrt{\mathrm{cot}{x}}{dx}\: \\ $$ Commented by Kunal12588 last updated on 16/May/20 $$=\int\sqrt{{tan}\left(\pi/\mathrm{2}−{x}\right)}\:{dx} \\ $$$$=−\int\sqrt{{tan}\:{t}}\:{dt}\:;\:{with}\:{t}={tan}\left(\frac{\pi}{\mathrm{2}}−{x}\right) \\ $$$$\int\sqrt{{tan}\:{x}}\:{dx}\:{is}\:{very}\:{popular}\:{integral},\:{its}\:{already} \\…

Question-159612

Question Number 159612 by cortano last updated on 19/Nov/21 Answered by Ar Brandon last updated on 19/Nov/21 $${I}=\int\frac{\sqrt{\mathrm{2}+\sqrt[{\mathrm{3}}]{{x}}}}{\:\sqrt[{\mathrm{3}}]{{x}}}{dx},\:{x}={u}^{\mathrm{3}} \Rightarrow{dx}=\mathrm{3}{u}^{\mathrm{2}} {du} \\ $$$$\:\:\:=\mathrm{3}\int\frac{\sqrt{\mathrm{2}+{u}}}{{u}}\centerdot{u}^{\mathrm{2}} {du}=\mathrm{3}\int{u}\sqrt{\mathrm{2}+{u}}{du} \\ $$$$\:\:\:=\mathrm{3}\int\left({u}+\mathrm{2}\right)^{\frac{\mathrm{3}}{\mathrm{2}}}…