Question Number 105781 by IE last updated on 31/Jul/20 $$\mathrm{Please},\:\mathrm{I}\:\mathrm{need}\:\mathrm{help}. \\ $$$$\mathrm{Exercise} \\ $$$$\mathrm{We}\:\mathrm{have}\:: \\ $$$$\mathrm{J}_{\mathrm{n}} =\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \mathrm{tan}^{\mathrm{n}} \left(\mathrm{x}\right)\:\mathrm{dx} \\ $$$$ \\ $$$$\left.\mathrm{1}\right)\:\mathrm{Establish}\:\mathrm{a}\:\mathrm{recurrence}\:\mathrm{relation} \\…
Question Number 171301 by mpakhrur last updated on 12/Jun/22 $$\int_{\mathrm{1}} ^{\mathrm{2}} \mathrm{6}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{3} \\ $$ Commented by Kalebwizeman last updated on 12/Jun/22 $$\left.\frac{\mathrm{6}{x}^{\mathrm{3}} }{\mathrm{3}}−\frac{\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{3}{x}\right]\underset{\mathrm{1}}…
Question Number 105755 by bemath last updated on 31/Jul/20 $$\int\:\frac{{dx}}{\:\sqrt{{x}\sqrt{{x}}\:−{x}^{\mathrm{2}} }}\:? \\ $$ Answered by john santu last updated on 31/Jul/20 $${let}\:{u}\:=\:\sqrt{{x}}\:\Rightarrow\:\int\:\frac{\mathrm{2}{u}\:{du}}{\:\sqrt{{u}^{\mathrm{3}} −{u}^{\mathrm{4}} }}\:=\: \\…
Question Number 105753 by bramlex last updated on 31/Jul/20 $$\int\:\frac{{x}^{\mathrm{2}} +\mathrm{sin}\:^{\mathrm{2}} {x}}{{x}^{\mathrm{2}} +\mathrm{cos}\:^{\mathrm{2}} {x}}\:{dx}\: \\ $$ Commented by Ar Brandon last updated on 31/Jul/20 $$\mathrm{No}\:\mathrm{limits}\:?…
Question Number 105743 by Skabetix last updated on 31/Jul/20 Commented by Skabetix last updated on 31/Jul/20 $${thank}\:{you}\:{in}\:{advance} \\ $$ Answered by john santu last updated…
Question Number 171260 by udaythool last updated on 11/Jun/22 $$\underline{{Change}\:{to}\:{polar}\:{coordinates}:} \\ $$$$\underset{\mathrm{0}} {\int}^{\:\:\mathrm{4}{a}} \underset{{y}^{\mathrm{2}} /\mathrm{4}{a}} {\int}\overset{{a}} {\:}\:\:\left(\frac{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\right)\:{dx}\:{dy} \\ $$ Answered by…
Question Number 105722 by bemath last updated on 31/Jul/20 $$\int\:\sqrt{{x}−\sqrt{{x}}}\:{dx}\: \\ $$ Commented by bemath last updated on 31/Jul/20 $${thank}\:{you}\:{all}\: \\ $$ Answered by bobhans…
Question Number 171253 by vonem1 last updated on 11/Jun/22 Answered by haladu last updated on 11/Jun/22 $$\:\mathrm{8}\int_{\mathrm{1}} ^{\mathrm{4}} \:\boldsymbol{\mathrm{t}}^{−\frac{\mathrm{1}}{\mathrm{2}}} \:\:\:\:\boldsymbol{\mathrm{dt}}\:−\mathrm{12}\:\:\int\:\boldsymbol{\mathrm{t}}^{\frac{\mathrm{3}}{\mathrm{2}}} \:\:\boldsymbol{\mathrm{dt}} \\ $$$$\:\:\: \\ $$$$\:\:\mathrm{8}\:\:\:\frac{\boldsymbol{\mathrm{t}}^{−\frac{\mathrm{1}}{\mathrm{2}}\:+\mathrm{1}}…
Question Number 105700 by john santu last updated on 31/Jul/20 $$\underset{−\pi} {\overset{\pi} {\int}}\:\frac{{x}^{\mathrm{2}} \:{dx}}{\mathrm{1}+\mathrm{sin}\:\left(\mathrm{sin}\:{x}\right)+\sqrt{\mathrm{1}+\mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{sin}\:{x}\right)}} \\ $$$$ \\ $$ Answered by bramlex last updated on…
Question Number 40161 by maxmathsup by imad last updated on 16/Jul/18 $${study}\:{the}\:{convergence}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{sin}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}{{ln}\left(\mathrm{1}+\sqrt{{x}}\right)}{dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com