Question Number 45795 by maxmathsup by imad last updated on 16/Oct/18 $${find}\:\int\:\frac{{dx}}{{cosx}\:{sin}^{\mathrm{2}} {x}} \\ $$ Answered by MJS last updated on 17/Oct/18 $$\frac{\mathrm{1}}{\mathrm{cos}\:{x}\:\mathrm{sin}^{\mathrm{2}} \:{x}}=\left(\mathrm{1}+\frac{\mathrm{cos}^{\mathrm{2}} \:{x}}{\mathrm{sin}^{\mathrm{2}}…
Question Number 45771 by maxmathsup by imad last updated on 16/Oct/18 $${find}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:{cos}\left({x}+{t}^{\mathrm{2}} \right){dtand}\:{g}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:{sin}\left({x}+{t}^{\mathrm{2}} \right){dt} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:{f}^{'} \left({x}\right)\:{and}\:{g}^{'} \left({x}\right). \\ $$ Answered…
Question Number 45735 by last updated on 16/Oct/18 $$\int_{\alpha} ^{\beta} \frac{\mathrm{1}}{\left({x}−\alpha\right)\left(\beta−{x}\right)}{dx}\:\:=?\:\:\:\beta>\alpha \\ $$ Commented by maxmathsup by imad last updated on 17/Oct/18 $${I}\:=−\int_{\alpha} ^{\beta}…
Question Number 45721 by Necxx last updated on 15/Oct/18 $$\:{Integrate}\:\mathrm{sin}\:\left({x}^{\mathrm{2}} \right){dx} \\ $$ Commented by MJS last updated on 16/Oct/18 $$\mathrm{not}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{resolve}\:\mathrm{to}\:\mathrm{elementary}\:\mathrm{function} \\ $$ Commented by…
Question Number 45706 by Meritguide1234 last updated on 15/Oct/18 Answered by tanmay.chaudhury50@gmail.com last updated on 17/Oct/18 $${trying}\:{to}\:{solve}… \\ $$$$\int\frac{\left(\mathrm{1}+{x}^{\mathrm{4}} \right)}{\left(\mathrm{1}−{x}^{\mathrm{4}} \right)\sqrt{\mathrm{1}+{x}^{\mathrm{4}} }\:}{dx} \\ $$$$\int\frac{\mathrm{1}+{x}^{\mathrm{4}} }{{x}^{\mathrm{2}}…
Question Number 45705 by Sanjarbek last updated on 15/Oct/18 Commented by maxmathsup by imad last updated on 16/Oct/18 $$\int\:{sin}\left({x}^{\mathrm{2}} \right){dx}\:=\frac{\sqrt{\pi}\left(\sqrt{\mathrm{2}}+{i}\sqrt{\mathrm{2}}\right){erf}\left\{\:\left(\sqrt{\mathrm{2}}+{i}\sqrt{\mathrm{2}}\right)\frac{{x}}{\mathrm{2}}\right\}+\sqrt{\pi}\left(\sqrt{\mathrm{2}}−{i}\sqrt{\mathrm{2}}\right){erf}\left\{\left(\sqrt{\mathrm{2}−}{i}\sqrt{\mathrm{2}}\right)\frac{{x}}{\mathrm{2}}\right\}}{\mathrm{8}} \\ $$$${this}\:{formulae}\:{is}\:{given}\:{by}\:{integral}\:{calculator}\:{so}\:{give}\:{me}\:{time}\:{to}\:{prof}\:{this}… \\ $$ Commented…
Question Number 45670 by arvinddayama01@gmail.com last updated on 15/Oct/18 $$\int{cos}^{−\mathrm{1}} \left({sinx}\right){dx}=? \\ $$ Commented by maxmathsup by imad last updated on 15/Oct/18 $${let}\:{I}\:=\int\:{arccos}\left({sinx}\right){dx}\:\:{changement}\:{arcos}\left({sinx}\right)={t}\:\Rightarrow{sinx}={cost} \\ $$$$\Rightarrow{x}={arcsin}\left({cost}\right)\:\Rightarrow{dx}=−{sint}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{cos}^{\mathrm{2}}…
Question Number 45669 by arvinddayama01@gmail.com last updated on 15/Oct/18 $$\int{tan}^{−\mathrm{1}} \sqrt{\frac{\mathrm{1}−{sinx}}{\mathrm{1}+{sinx}}}\:{dx}=? \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 15/Oct/18 $${t}={tan}\frac{{x}}{\mathrm{2}}\:\:\: \\ $$$$ \\ $$$${now}\:{tan}^{−\mathrm{1}}…
Question Number 111195 by bemath last updated on 02/Sep/20 $$\:\:\:\:\sqrt{\mathrm{bemath}} \\ $$$$\:\:\:\int\:\frac{\mathrm{dx}}{\:\sqrt[{\mathrm{3}\:}]{\mathrm{x}−\mathrm{1}}\:\:\sqrt[{\mathrm{3}\:}]{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }}\:? \\ $$ Answered by Sarah85 last updated on 02/Sep/20 $$\mathrm{substitute}\:{t}=\sqrt[{\mathrm{3}}]{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}}\:\Leftrightarrow\:{x}=−\frac{{t}^{\mathrm{3}} +\mathrm{1}}{{t}^{\mathrm{3}} −\mathrm{1}}\:\Rightarrow…
Question Number 111189 by john santu last updated on 02/Sep/20 $$\:\left(\mathrm{1}\right)\:\:\:\:\:\:\:\int\:\frac{\left({x}+\mathrm{1}\right){dx}}{{x}^{\mathrm{4}} \left({x}−\mathrm{1}\right)}\:?\: \\ $$$$\:\left(\mathrm{2}\right)\:\:\:\:\:\:\frac{{dy}}{{dx}}\:+\:\frac{{y}}{{x}−\mathrm{2}}\:=\:\mathrm{5}\left({x}−\mathrm{2}\right)\sqrt{{y}}\: \\ $$ Answered by Sarah85 last updated on 02/Sep/20 $$\frac{{x}+\mathrm{1}}{{x}^{\mathrm{4}} \left({x}−\mathrm{1}\right)}=\frac{\mathrm{2}}{{x}−\mathrm{1}}−\frac{\mathrm{2}}{{x}}−\frac{\mathrm{2}}{{x}^{\mathrm{2}}…