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Category: Integration

Question-84708

Question Number 84708 by Power last updated on 15/Mar/20 Answered by TANMAY PANACEA last updated on 16/Mar/20 $$\int\frac{{dx}}{\left({x}+\mathrm{2}\right)^{\mathrm{2}} \sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{5}}} \\ $$$$\int\frac{{dx}}{\left({x}+\mathrm{2}\right)^{\mathrm{2}} \sqrt{\left({x}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{6}}} \\…

Question-84702

Question Number 84702 by Power last updated on 15/Mar/20 Commented by abdomathmax last updated on 15/Mar/20 $${I}\:=\int\:\:\:\frac{{dx}}{{shx}+\mathrm{1}}\:\Rightarrow{I}=\int\:\:\frac{{dx}}{\frac{{e}^{{x}} −{e}^{−{x}} }{\mathrm{2}}+\mathrm{1}}\:=\int\frac{\mathrm{2}{dx}}{{e}^{{x}} −{e}^{−{x}} \:+\mathrm{2}} \\ $$$$=_{{e}^{{x}} ={t}} \:\:\:\:\:\int\:\:\frac{\mathrm{2}}{{t}−{t}^{−\mathrm{1}}…

tan-x-1-3-dx-

Question Number 84709 by M±th+et£s last updated on 15/Mar/20 $$\int\sqrt[{\mathrm{3}}]{{tan}\left({x}\right)}\:{dx} \\ $$ Answered by MJS last updated on 15/Mar/20 $$\int\sqrt[{\mathrm{3}}]{\mathrm{tan}\:{x}}\:{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\left(\mathrm{tan}\:{x}\right)^{\mathrm{2}/\mathrm{3}} \:\rightarrow\:{dx}=\frac{\mathrm{3}}{\mathrm{2}}\left(\mathrm{sin}\:{x}\right)^{\mathrm{1}/\mathrm{3}} \left(\mathrm{cos}\:{x}\right)^{\mathrm{5}/\mathrm{3}} {dt}\right]…

Question-84693

Question Number 84693 by Power last updated on 15/Mar/20 Commented by abdomathmax last updated on 15/Mar/20 $${let}\:\varphi\left({t}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{{sin}\left({tx}\right)}{{x}\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)}{dx}\:\:{we}\:{have} \\ $$$$\varphi^{'} \left({t}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{{cos}\left({tx}\right)}{{x}^{\mathrm{2}}…

Question-84681

Question Number 84681 by Power last updated on 15/Mar/20 Commented by mathmax by abdo last updated on 15/Mar/20 $${I}\:=\int\:\:\frac{\mathrm{15}{sinx}\:+\mathrm{2}{cosx}}{\mathrm{98}{sinx}\:−\mathrm{7}{cosx}}{dx}\:\Rightarrow{I}\:=\frac{\mathrm{15}}{\mathrm{98}}\int\:\:\frac{{sinx}+\frac{\mathrm{2}}{\mathrm{15}}{cosx}}{{sinx}−\frac{\mathrm{7}}{\mathrm{98}}{cosx}}{dx} \\ $$$${let}\:{determine}\:{A}\:=\int\:\:\frac{{sinx}+{acosx}}{{sinx}\:+{bcosx}}{dx}\:\:{we}\:{di}\:{the}\:{changement} \\ $$$${tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:\Rightarrow\:{A}\:=\int\frac{\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }+{a}\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}}…