Question Number 167647 by mathlove last updated on 22/Mar/22 Answered by RoswelCod2003 last updated on 22/Mar/22 $$\mathrm{Using}\:“\mathrm{Feynman}'\mathrm{s}\:\mathrm{Rule}\:\mathrm{of}\:\mathrm{Integration}'' \\ $$$$\: \\ $$$${I}\left({k}\right)\:=\:\int_{\mathrm{0}} ^{\:\infty} {e}^{−{x}^{\mathrm{2}} } \mathrm{cos}\:\left({kx}\right)\:{dx}\:\therefore\:\mathrm{where}\:{k}\:=\:\mathrm{2}…
Question Number 102099 by Study last updated on 06/Jul/20 $$\int{sinx}\:\centerdot\:{cosx}\:\centerdot{cos}\mathrm{2}{x}\:\centerdot\:{cos}\mathrm{4}{x}\:{dx}=? \\ $$ Answered by PRITHWISH SEN 2 last updated on 06/Jul/20 $$\frac{\mathrm{1}}{\mathrm{2}}\int\mathrm{sin}\:\mathrm{2xcos}\:\mathrm{2xcos}\:\mathrm{4x}\:\mathrm{dx}=\frac{\mathrm{1}}{\mathrm{4}}\int\mathrm{sin}\:\mathrm{4xcos}\:\mathrm{4xdx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\int\mathrm{sin}\:\:\mathrm{8xdx}=−\frac{\mathrm{cos}\:\:\mathrm{8x}}{\mathrm{64}}\:+\mathrm{C} \\…
Question Number 102097 by Study last updated on 06/Jul/20 $$\int\frac{{cos}^{\mathrm{2}} {x}\:−{cos}^{\mathrm{2}} {x}}{{cosx}−{cosx}}{dx}=? \\ $$ Commented by Dwaipayan Shikari last updated on 06/Jul/20 $$\mathrm{2}{sinx} \\ $$…
Question Number 167626 by dangduomg last updated on 21/Mar/22 $$\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{t}} }{\:\sqrt{{t}}}\:{e}^{−\frac{\mathrm{1}}{\mathrm{4}{t}}} \:{dt} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 36547 by tanmay.chaudhury50@gmail.com last updated on 03/Jun/18 $$\int\frac{\mathrm{5}{x}^{\mathrm{4}} +\mathrm{4}{x}^{\mathrm{5}} }{\left({x}^{\mathrm{5}} +{x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$ Answered by ajfour last updated on 03/Jun/18 $$=−\int\frac{\left(−\frac{\mathrm{5}}{{x}^{\mathrm{6}} }−\frac{\mathrm{4}}{{x}^{\mathrm{5}}…
Question Number 36545 by tanmay.chaudhury50@gmail.com last updated on 03/Jun/18 $$\int\frac{\mathrm{2}{x}+\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{dx} \\ $$ Commented by MJS last updated on 03/Jun/18 $$…\mathrm{I}\:\mathrm{just}\:\mathrm{gave}\:\mathrm{it}\:\mathrm{a}\:\mathrm{try}\:\mathrm{because}\:\mathrm{the}\:“\mathrm{poper}''\: \\ $$$$\mathrm{way}\:\mathrm{looked}\:\mathrm{rather}\:\mathrm{complicated}… \\…
Question Number 36546 by tanmay.chaudhury50@gmail.com last updated on 03/Jun/18 $$\int\frac{\left({x}−\mathrm{1}\right)}{\left({x}+\mathrm{1}\right)\sqrt{{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}}}{dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 36544 by tanmay.chaudhury50@gmail.com last updated on 03/Jun/18 $$\int\frac{{dx}}{{tanx}+{cotx}+{secx}+{cosecx}} \\ $$ Answered by ajfour last updated on 03/Jun/18 $${I}=\int\frac{\mathrm{sin}\:{x}\mathrm{cos}\:{xdx}}{\mathrm{1}+\mathrm{sin}\:{x}+\mathrm{cos}\:{x}} \\ $$$$\:\:\:\:=\int\frac{\mathrm{2sin}\:\frac{{x}}{\mathrm{2}}\mathrm{cos}\:\frac{{x}}{\mathrm{2}}\left(\mathrm{cos}\:^{\mathrm{2}} \frac{{x}}{\mathrm{2}}−\mathrm{sin}\:^{\mathrm{2}} \frac{{x}}{\mathrm{2}}\right)}{\mathrm{2cos}\:\frac{{x}}{\mathrm{2}}\left(\mathrm{sin}\:\frac{{x}}{\mathrm{2}}+\mathrm{cos}\:\frac{{x}}{\mathrm{2}}\right)}{dx} \\…
Question Number 36543 by tanmay.chaudhury50@gmail.com last updated on 03/Jun/18 $$\int{cot}^{−\mathrm{1}} \left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\:{dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 36538 by tanmay.chaudhury50@gmail.com last updated on 03/Jun/18 $$\int\frac{{dx}}{\left({a}+{bx}^{\mathrm{2}} \right)\sqrt{{b}−{ax}^{\mathrm{2}} \:}}\:\: \\ $$ Commented by abdo.msup.com last updated on 03/Jun/18 $${we}\:{have}\:{b}\:−{ax}^{\mathrm{2}} ={b}\left(\mathrm{1}−\left(\sqrt{\frac{{a}}{{b}}}\:{x}\right)^{\mathrm{2}} \:\:{if}\:\frac{{a}}{{b}}>\mathrm{0}\right. \\…