Question Number 103241 by Dwaipayan Shikari last updated on 13/Jul/20 $$\int{x}^{−{x}} {dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 168772 by Dildora last updated on 17/Apr/22 Commented by safojontoshtemirov last updated on 18/Apr/22 $${S}=\mathrm{4}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\sqrt{\left(\left({e}^{{t}} {sint}\right)'\right)^{\mathrm{2}} +\left(\left({e}^{{t}} {cost}\right)'\right)^{\mathrm{2}} }{dt}\: \\ $$$${S}=\mathrm{4}\underset{\mathrm{0}}…
Question Number 168771 by Dildora last updated on 17/Apr/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 168762 by Dildora last updated on 17/Apr/22 Commented by cortano1 last updated on 17/Apr/22 $$\:\int\:\frac{\mathrm{2sin}\:\frac{\mathrm{1}}{\mathrm{2}}{x}\:\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}{x}}{\mathrm{2cos}\:\frac{\mathrm{1}}{\mathrm{2}}{x}\left(\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}{x}+\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}{x}\right)}\:{dx} \\ $$$$\:=\:\int\:\frac{\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}{x}}{\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}{x}+\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}{x}}\:{dx} \\ $$$$\:=\:\int\:\frac{\mathrm{2sin}\:{u}}{\mathrm{cos}\:{u}+\mathrm{sin}\:{u}}\:{du}\:;\:\left[\frac{\mathrm{1}}{\mathrm{2}}{x}={u}\:\right] \\ $$$$\frac{\mathrm{2sin}\:{u}}{\mathrm{cos}\:{u}+\mathrm{sin}\:{u}}\:=\:{A}\left(\frac{\mathrm{sin}\:{u}+\mathrm{cos}\:{u}}{\mathrm{sin}\:{u}+\mathrm{cos}\:{u}}\right)+{B}\left(\frac{{d}\left(\mathrm{sin}\:{u}+\mathrm{cos}\:{u}\right)}{\mathrm{sin}\:{u}+\mathrm{cos}\:{u}}\right) \\ $$$$\Rightarrow\mathrm{2sin}\:{u}=\:{A}\mathrm{sin}\:{u}+{A}\mathrm{cos}\:{u}+{B}\mathrm{cos}\:\:{u}−{B}\mathrm{sin}\:{u}…
Question Number 103220 by Dwaipayan Shikari last updated on 13/Jul/20 $$\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{3}} }{{e}^{{x}} +\mathrm{1}}{dx} \\ $$ Answered by mathmax by abdo last updated on…
Question Number 168745 by MikeH last updated on 17/Apr/22 $$\int\frac{\mathrm{1}}{{x}+\sqrt{{x}−\mathrm{1}}}\:{dx}\:=\:?? \\ $$ Commented by safojontoshtemirov last updated on 17/Apr/22 $$\int\frac{{dx}}{{x}+\sqrt{{x}−\mathrm{1}}}=\:\:\sqrt{{x}−\mathrm{1}}={t}\:\:;\:{x}={t}^{\mathrm{2}} +\mathrm{1}\:\:;{dx}=\mathrm{2}{tdt} \\ $$$$\int\frac{\mathrm{2}{tdt}}{{t}^{\mathrm{2}} +{t}+\mathrm{1}}=\int\frac{\mathrm{2}{t}+\mathrm{1}}{{t}^{\mathrm{2}} +{t}+\mathrm{1}}{dt}−\int\frac{\mathrm{1}}{{t}^{\mathrm{2}}…
Question Number 168743 by mehdiAz last updated on 16/Apr/22 $$\int_{\mathrm{0}} ^{\infty} \frac{\sqrt{{t}}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$\mathrm{FAILED}\:\mathrm{TO}\:\mathrm{CALCULATE} \\ $$$$ \\ $$ Commented by GalaxyBills last updated on…
Question Number 103198 by Dwaipayan Shikari last updated on 13/Jul/20 $$\int_{\mathrm{0}} ^{\mathrm{1}} {sin}\left({logx}\right){dx} \\ $$ Commented by Dwaipayan Shikari last updated on 13/Jul/20 $${sinx}=\frac{{e}^{{ix}} −{e}^{−{ix}}…
Question Number 168732 by mnjuly1970 last updated on 16/Apr/22 Answered by mnjuly1970 last updated on 16/Apr/22 $$\:\:\:\:\:\:\:\:\Omega\:=\:\int_{\mathrm{0}} ^{\:\infty} \frac{\:{x}^{\:\frac{\mathrm{1}}{\mathrm{4}}} +\:\mathrm{2}{x}^{\:\:\frac{\mathrm{5}}{\mathrm{4}}} +{x}^{\:\frac{\mathrm{9}}{\mathrm{4}}} }{\left(\mathrm{1}+{x}^{\:\mathrm{3}} \right)^{\:\mathrm{2}} }{dx} \\…
Question Number 103196 by bobhans last updated on 13/Jul/20 $$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}−{x}\right)\:\mathrm{ln}\left(\mathrm{1}−{x}\right)\:{dx}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}\:? \\ $$ Answered by bramlex last updated on 13/Jul/20 $$\mathrm{replace}\:\mathrm{x}\:\mathrm{with}\:\mathrm{1}−\mathrm{x}\: \\ $$$$\mathrm{I}\:=\:\underset{\mathrm{0}}…