Question Number 85395 by Power last updated on 21/Mar/20 Commented by MJS last updated on 21/Mar/20 $$\mathrm{let}\:{t}=\sqrt{\mathrm{ctg}\:{x}}\:\rightarrow\:{dx}=−\mathrm{2sin}^{\mathrm{2}} \:{x}\:\sqrt{\mathrm{ctg}\:{x}} \\ $$$$\Rightarrow\:\int\sqrt{\mathrm{ctg}\:{x}}{dx}=−\mathrm{2}\int\frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{4}} +\mathrm{1}}{dt} \\ $$$$\mathrm{and}\:\mathrm{this}\:\mathrm{should}\:\mathrm{be}\:\mathrm{easy} \\…
Question Number 85391 by sakeefhasan05@gmail.com last updated on 21/Mar/20 Commented by sakeefhasan05@gmail.com last updated on 21/Mar/20 $$\mathrm{I}\:\mathrm{have}\:\mathrm{got}\:\left(\frac{\mathrm{8}}{\mathrm{63}}\right)\:\mathrm{as}\:\mathrm{answer}\: \\ $$$$\mathrm{whoever}\:\mathrm{check}\:\mathrm{it}\:\mathrm{pls}\: \\ $$ Commented by mathmax by…
Question Number 85384 by sakeefhasan05@gmail.com last updated on 21/Mar/20 Commented by mathmax by abdo last updated on 21/Mar/20 $${let}\:{A}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{4}} }\:\Rightarrow\:{A}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}}…
Question Number 85374 by M±th+et£s last updated on 21/Mar/20 $$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{7}} +{x}^{\mathrm{3}} +\mathrm{1}}{{x}^{\mathrm{4}} +\mathrm{1}}\:{dx}\: \\ $$ Commented by MJS last updated on 21/Mar/20 $$\frac{{x}^{\mathrm{7}}…
Question Number 85383 by M±th+et£s last updated on 21/Mar/20 $$\int\sqrt[{\mathrm{5}}]{\frac{{sin}\left({x}\right)}{{cos}^{\mathrm{11}} \left({x}\right)}}\:{dx} \\ $$ Answered by john santu last updated on 21/Mar/20 $$\int\:\:\:\frac{\sqrt[{\mathrm{5}\:}]{\mathrm{tan}\:{x}}}{\mathrm{cos}\:^{\mathrm{2}} {x}}\:{dx}\:=\:\int\:\:\sqrt[{\mathrm{5}\:\:}]{\mathrm{tan}\:{x}}\:{d}\left(\mathrm{tan}\:{x}\right) \\ $$$$=\:\frac{\mathrm{5}}{\mathrm{6}}\:\sqrt[{\mathrm{5}\:\:}]{\left(\mathrm{tan}\:{x}\right)^{\mathrm{6}}…
Question Number 85362 by sahnaz last updated on 21/Mar/20 $$\int\frac{−\mathrm{4}−\mathrm{u}}{\mathrm{u}^{\mathrm{2}} +\mathrm{5u}+\mathrm{6}}\mathrm{du} \\ $$ Answered by john santu last updated on 22/Mar/20 $$\int\:\frac{{du}}{{u}+\mathrm{3}}\:−\int\:\frac{\mathrm{2}}{{u}+\mathrm{2}}\:{du}\:=\: \\ $$$$\:\mathrm{ln}\:\mid{u}+\mathrm{3}\mid\:−\:\mathrm{2}\:\mathrm{ln}\:\mid{u}+\mathrm{2}\mid\:+\:{c}\: \\…
Question Number 85360 by sahnaz last updated on 21/Mar/20 $$\int\left(\mathrm{e}^{\left(\mathrm{1}−\mathrm{x}\right)×\mathrm{e}^{\mathrm{x}} } ×\mathrm{e}^{\int\mathrm{xe}^{\mathrm{x}} \mathrm{dx}} \right)\mathrm{dx} \\ $$ Answered by john santu last updated on 21/Mar/20 $$\int\:{xe}^{{x}}…
Question Number 85355 by TawaTawa1 last updated on 21/Mar/20 Commented by mathmax by abdo last updated on 21/Mar/20 $${I}\:=\int\:{arctan}\left(\sqrt{\mathrm{1}+\sqrt{{x}}}\right){dx}\:\:{changement}\:\sqrt{\mathrm{1}+\sqrt{{x}}}={t}\:{give} \\ $$$$\mathrm{1}+\sqrt{{x}}={t}^{\mathrm{2}} \:\Rightarrow\sqrt{{x}}={t}^{\mathrm{2}} −\mathrm{1}\:\Rightarrow{x}\:=\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} \:\Rightarrow{dx}\:=\mathrm{2}\left(\mathrm{2}{t}\right)\left({t}^{\mathrm{2}}…
Question Number 150871 by liberty last updated on 16/Aug/21 $$\:\mathrm{Find}\:\mathrm{all}\:\mathrm{the}\:\mathrm{real}\:\mathrm{solutions} \\ $$$$\mathrm{of}\:\mathrm{cos}\:\mathrm{x}+\mathrm{cos}\:^{\mathrm{5}} \mathrm{x}+\mathrm{cos}\:\mathrm{7x}=\mathrm{3} \\ $$ Answered by EDWIN88 last updated on 16/Aug/21 $$\:\mathrm{cos}\:\mathrm{7}{x}+\mathrm{cos}\:{x}=\mathrm{2cos}\:\mathrm{4}{x}\:\mathrm{cos}\:\mathrm{3}{x} \\ $$$$\:=\:\mathrm{2}\left(\mathrm{2cos}\:^{\mathrm{2}}…
Question Number 85323 by sahnaz last updated on 20/Mar/20 $$\int\frac{\mathrm{z}−\mathrm{3}}{\mathrm{5z}−\mathrm{10}}\mathrm{dz} \\ $$ Answered by MJS last updated on 20/Mar/20 $$\int\frac{{z}−\mathrm{3}}{\mathrm{5}{z}−\mathrm{10}}{dz}=\frac{\mathrm{1}}{\mathrm{5}}\int\frac{{z}−\mathrm{3}}{{z}−\mathrm{2}}{dz}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}}\int{dz}−\frac{\mathrm{1}}{\mathrm{5}}\int\frac{{dz}}{{z}−\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{5}}{z}−\frac{\mathrm{1}}{\mathrm{5}}\mathrm{ln}\:\mid{z}−\mathrm{2}\mid\:+{C} \\ $$ Terms…