Question Number 41634 by Tawa1 last updated on 10/Aug/18 Commented by maxmathsup by imad last updated on 10/Aug/18 $${I}\:\:=\:\int\:\:\:\:\frac{{dx}}{{sinx}\:+\frac{\mathrm{1}}{{cosx}}}\:=\:\:\int\:\:\:\:\:\:\frac{{cosx}}{{sinx}\:.{cosx}\:+\mathrm{1}}{dx}\:\:{changement}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:{give} \\ $$$${I}\:=\:\int\:\:\:\:\:\:\frac{\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}{\frac{\mathrm{2}{t}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}}…
Question Number 41627 by alex041103 last updated on 10/Aug/18 Commented by alex041103 last updated on 10/Aug/18 $${Just}\:{for}\:{fun}!\:{I}\:{have}\:{already}\:{posted} \\ $$$${this}\:{Q}.\:{around}\:{a}\:{year}\:{ago}.\:{The}\:{result} \\ $$$${is}\:{beautiful}\:{and}\:{worths}\:{giving}\:{it}\:{a}\:{try}. \\ $$$${Can}\:{you}\:{prove}\:{something}\:{interesting}. \\ $$$${Hint}:\:{Calculate}\:{the}\:{result}.…
Question Number 107162 by bobhans last updated on 09/Aug/20 Commented by Dwaipayan Shikari last updated on 09/Aug/20 $$\int_{−\mathrm{5}} ^{\mathrm{5}} \sqrt[{\mathrm{6}}]{\mathrm{5x}^{\mathrm{2}} +\mathrm{6}}\:\left(\mathrm{5x}−\mathrm{2}\right)\mathrm{dx}=\int_{−\mathrm{5}} ^{\mathrm{5}} \sqrt[{\mathrm{6}}]{\mathrm{5x}^{\mathrm{2}} +\mathrm{6}}\:\left(−\mathrm{5x}−\mathrm{2}\right)=\mathrm{I} \\…
Question Number 107141 by cesarL last updated on 09/Aug/20 $$\int\left({x}^{\mathrm{3}} +{x}^{\mathrm{6}} \right)\left(\sqrt[{\mathrm{3}}]{{x}^{\mathrm{3}} +\mathrm{2}}\right){dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 41586 by MJS last updated on 09/Aug/18 $${f}\left({x}\right)=\sqrt{−\mathrm{3}+\sqrt{\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}}} \\ $$$$\int{f}\left({x}\right)=? \\ $$$$\int{f}^{−\mathrm{1}} \left({x}\right)=? \\ $$ Commented by prof Abdo imad last updated on…
Question Number 172653 by Mikenice last updated on 29/Jun/22 Answered by MJS_new last updated on 30/Jun/22 $$\int\frac{\left(−\mathrm{1}\right)^{\mathrm{1}/{x}} }{{x}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{\pi\mathrm{i}}{{x}}\:\rightarrow\:{dx}=\frac{\mathrm{i}{x}^{\mathrm{2}} }{\pi}{dt}\right] \\ $$$$=−\int\frac{\mathrm{e}^{{t}} }{{t}}{dt}=−\mathrm{Ei}\:\left({t}\right)\:= \\…
Question Number 107113 by cesarL last updated on 08/Aug/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 172651 by Mikenice last updated on 29/Jun/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 107107 by mohammad17 last updated on 08/Aug/20 Commented by Dwaipayan Shikari last updated on 08/Aug/20 $$\int_{\mathrm{0}} ^{\infty} \sqrt{\mathrm{y}}\:\:\mathrm{e}^{−\mathrm{t}^{\mathrm{2}} } \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{y}^{\mathrm{3}} =\mathrm{t}^{\mathrm{2}\:\:} \:,\mathrm{y}^{\frac{\mathrm{3}}{\mathrm{2}}} =\mathrm{t}\:\:\:\:\:\frac{\mathrm{3}}{\mathrm{2}}\sqrt{\mathrm{y}}\:\:=\frac{\mathrm{dt}}{\mathrm{dy}}…
Question Number 172636 by Mikenice last updated on 29/Jun/22 Answered by floor(10²Eta[1]) last updated on 29/Jun/22 $$\mathrm{I}=\int\frac{\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}−\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}}\mathrm{dx}=\int\frac{\mathrm{dx}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}}−\int\frac{\mathrm{dx}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}}=\mathrm{I}_{\mathrm{1}} −\mathrm{I}_{\mathrm{2}} \\…