Question Number 196817 by universe last updated on 01/Sep/23 Answered by witcher3 last updated on 04/Sep/23 $$\mathrm{are}\:\mathrm{You}\:\mathrm{sur}\:\mathrm{of}\:\mathrm{the}\:\mathrm{expression}? \\ $$ Commented by universe last updated on…
Question Number 196788 by ERLY last updated on 31/Aug/23 $${calculer}\:\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$<{erly}\:{rolvinst}> \\ $$$$ \\ $$ Answered by MrGHK last updated on…
Question Number 196688 by cortano12 last updated on 29/Aug/23 Answered by Frix last updated on 29/Aug/23 $$\int\sqrt{{x}−\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}}\:{dx}\:\overset{{t}=\sqrt{{x}−\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}}} {=} \\ $$$$=\int\frac{{t}^{\mathrm{4}} −\mathrm{4}}{{t}^{\mathrm{2}} }{dt}=\frac{{t}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{4}}{{t}}=…
Question Number 196557 by BHOOPENDRA last updated on 27/Aug/23 Answered by qaz last updated on 27/Aug/23 $$\int_{\mathrm{0}} ^{{t}} {e}^{−{u}} \mathrm{sin}\:{udu}=−\Im\int_{\mathrm{0}} ^{{t}} {e}^{−\left(\mathrm{1}+{i}\right){u}} {du}=\Im\frac{\mathrm{1}}{\mathrm{1}+{i}}\left({e}^{−\left(\mathrm{1}+{i}\right){t}} −\mathrm{1}\right) \\…
Question Number 196496 by RoseAli last updated on 26/Aug/23 Answered by witcher3 last updated on 26/Aug/23 $$\int\frac{\mathrm{1}}{\mathrm{tg}^{\mathrm{2}} \left(\mathrm{x}\right).\mathrm{cos}^{\mathrm{6}} \left(\mathrm{x}\right)} \\ $$$$=\int\frac{\mathrm{1}+\mathrm{tg}^{\mathrm{2}} \left(\mathrm{x}\right)}{\mathrm{tg}^{\mathrm{2}} \left(\mathrm{x}\right)}.\left(\mathrm{1}+\mathrm{tg}^{\mathrm{2}} \left(\mathrm{x}\right)\right)^{\mathrm{2}} \mathrm{dx}…
Question Number 196459 by RoseAli last updated on 25/Aug/23 Answered by Frix last updated on 25/Aug/23 $$\mathrm{Use}\:\mathrm{Ostrogradski}'\mathrm{s}\:\mathrm{Method}\:\mathrm{to}\:\mathrm{get} \\ $$$$\int\frac{{x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }{dx}=\frac{{x}\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}\right)}{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{{x}^{\mathrm{2}}…
Question Number 196487 by RoseAli last updated on 25/Aug/23 $$\int\left(\mathrm{sec}\:^{\mathrm{4}} {x}−\mathrm{cot}\:^{\mathrm{4}} {x}\right){dx} \\ $$ Answered by MM42 last updated on 26/Aug/23 $${I}_{\mathrm{1}} =\int{sec}^{\mathrm{4}} {xdx}=\int\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right)\left(\mathrm{1}+{tan}^{\mathrm{2}}…
Question Number 196436 by RoseAli last updated on 24/Aug/23 $$\int\frac{{dx}}{{x}\left({x}^{{n}} −\mathrm{1}\right)} \\ $$ Answered by MM42 last updated on 24/Aug/23 $$\int\frac{{x}^{{n}} −\mathrm{1}+{x}^{{n}} }{{x}\left({x}^{{n}} −\mathrm{1}\right)}=\int\left(\frac{\mathrm{1}}{{x}}+\frac{{x}^{{n}−\mathrm{1}} }{{x}^{{n}}…
Question Number 196435 by RoseAli last updated on 24/Aug/23 Commented by Denis last updated on 24/Aug/23 Commented by RoseAli last updated on 24/Aug/23 $$\boldsymbol{\mathrm{thanks}} \\…
Question Number 196375 by RoseAli last updated on 23/Aug/23 $$\int\frac{{dx}}{{x}\left({x}^{\mathrm{4}} −\mathrm{1}\right)} \\ $$ Commented by RoseAli last updated on 23/Aug/23 $$\int\frac{{dx}}{{x}\left({x}^{\mathrm{4}} +\mathrm{1}\right)} \\ $$ Answered…