Question Number 17463 by alex041103 last updated on 06/Jul/17 $${Evaluate}\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\:\frac{{x}^{\mathrm{4}} \left(\mathrm{1}−{x}\right)^{\mathrm{4}} }{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx}. \\ $$ Answered by ajfour last updated on 06/Jul/17 $$\frac{\mathrm{22}}{\mathrm{7}}−\pi\:.…
Question Number 17444 by Tinkutara last updated on 06/Jul/17 $$\mathrm{Evaluate}:\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}\frac{{dx}}{\mathrm{cos}^{\mathrm{3}} \:{x}\:\sqrt{\mathrm{2}\:\mathrm{sin}\:\mathrm{2}{x}}} \\ $$ Answered by Arnab Maiti last updated on 10/Jul/17 $$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}}…
Question Number 82974 by Power last updated on 26/Feb/20 Commented by Power last updated on 27/Feb/20 $$\mathrm{sir}\:\mathrm{mind}\:\mathrm{is}\:\mathrm{power}\:\:\mathrm{pls} \\ $$ Answered by mind is power last…
Question Number 82972 by mathmax by abdo last updated on 26/Feb/20 $${calculate}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{arctan}\left({x}\right){ln}\left(\mathrm{1}+{x}\right){dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 82970 by mathmax by abdo last updated on 26/Feb/20 $$\left.\mathrm{1}\right){find}\:\int\:\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{9}} } \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{2}} ^{+\infty} \:\frac{{dx}}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{9}} } \\ $$ Commented by mathmax…
Question Number 82971 by mathmax by abdo last updated on 26/Feb/20 $$\left.\mathrm{1}\right){find}\:\int\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{6}} } \\ $$$$\left.\mathrm{2}\right){calculate}\:\int_{−\infty} ^{+\infty} \:\frac{{dx}}{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{6}} } \\ $$ Commented by mathmax…
Question Number 82954 by john santu last updated on 26/Feb/20 $$\int\:\frac{\mathrm{cos}\:\mathrm{4x}−\mathrm{cos}\:\mathrm{2x}}{\mathrm{sin}\:\mathrm{4x}−\mathrm{cos}\:\mathrm{2x}}\:\mathrm{dx}\: \\ $$ Commented by john santu last updated on 26/Feb/20 $$\mathrm{let}\:\mathrm{u}=\:\mathrm{cos}\:\mathrm{2x}\:\Rightarrow\:\mathrm{dx}\:=\:−\frac{\mathrm{du}}{\mathrm{2}\sqrt{\mathrm{1}−\mathrm{u}^{\mathrm{2}} }} \\ $$$$\Rightarrow\int\:\frac{\:\mathrm{2u}^{\mathrm{2}}…
Question Number 148467 by bramlexs22 last updated on 28/Jul/21 Answered by puissant last updated on 28/Jul/21 $$\mathrm{x}=\mathrm{tan}\left(\mathrm{t}\right)\Rightarrow\mathrm{dx}=\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \left(\mathrm{t}\right)\mathrm{dt} \\ $$$$\mathrm{0}\leqslant\mathrm{x}\leqslant\infty\:\Rightarrow\:\mathrm{0}\leqslant\mathrm{t}\leqslant\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{arctan}\left(\mathrm{tan}\left(\mathrm{t}\right)\right)}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \left(\mathrm{t}\right)}\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}}…
Question Number 17392 by tawa tawa last updated on 05/Jul/17 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{cube}\:\mathrm{root}\:\mathrm{of}\:\:\mathrm{z}\:=\:−\:\mathrm{1} \\ $$ Answered by mrW1 last updated on 05/Jul/17 $$−\mathrm{1} \\ $$$$−\omega=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i} \\ $$$$−\omega^{\mathrm{2}}…
Question Number 17377 by tawa tawa last updated on 04/Jul/17 $$\int\:\:\frac{\mathrm{cos}\left(\mathrm{x}\right)}{\mathrm{2}\:−\:\mathrm{cos}\left(\mathrm{x}\right)}\:\mathrm{dx} \\ $$ Answered by ajfour last updated on 05/Jul/17 $$\mathrm{cos}\:\mathrm{x}=\frac{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\left(\mathrm{x}/\mathrm{2}\right)}{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \left(\mathrm{x}/\mathrm{2}\right)}\:=\frac{\mathrm{2}\left(\mathrm{1}−\mathrm{t}^{\mathrm{2}} \right)}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }…