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Category: Integration

Show-that-1-1-dx-5-cosh-x-13-sinh-x-1-2-log-e-15e-10-3e-2-

Question Number 29105 by tawa tawa last updated on 04/Feb/18 $$\mathrm{Show}\:\mathrm{that}:\:\:\:\int_{−\mathrm{1}} ^{\:\:\:\mathrm{1}} \:\:\:\:\:\:\:\frac{\mathrm{dx}}{\mathrm{5}\:\mathrm{cosh}\left(\mathrm{x}\right)\:+\:\mathrm{13}\:\mathrm{sinh}\left(\mathrm{x}\right)}\:\:=\:\:\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{log}_{\mathrm{e}} \left(\frac{\mathrm{15e}\:−\:\mathrm{10}}{\mathrm{3e}\:+\:\mathrm{2}}\right)\: \\ $$ Commented by abdo imad last updated on 04/Feb/18 $${I}=\:\int_{−\mathrm{1}}…

let-give-g-x-0-arctan-x-1-t-2-1-t-2-dt-find-a-simple-form-of-g-x-without-integral-

Question Number 29077 by abdo imad last updated on 04/Feb/18 $${let}\:{give}\:{g}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left({x}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:\:{find}\:{a}\:{simple} \\ $$$${form}\:{of}\:\:{g}^{'} \left({x}\right)\:{without}\:{integral}. \\ $$ Commented by abdo imad last…

let-give-f-x-0-1-arctan-x-1-t-2-1-t-2-dt-find-asimple-form-of-f-x-without-integral-

Question Number 29076 by abdo imad last updated on 04/Feb/18 $${let}\:{give}\:{f}\left({x}\right)=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{arctan}\left({x}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:\:{find}\:{asimple} \\ $$$${form}\:{of}\:{f}\left({x}\right)\:{without}\:{integral}. \\ $$ Commented by abdo imad last updated…

x-2-1-x-1-2x-3-dx-

Question Number 94609 by  M±th+et+s last updated on 20/May/20 $$\int\frac{{x}^{\mathrm{2}} −\mathrm{1}}{\:\sqrt{{x}+\mathrm{1}}+\sqrt{\mathrm{2}{x}+\mathrm{3}}}{dx} \\ $$ Answered by mathmax by abdo last updated on 20/May/20 $$\mathrm{I}\:=\int\:\:\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}{\:\sqrt{\mathrm{x}+\mathrm{1}}+\sqrt{\mathrm{2x}+\mathrm{3}}}\mathrm{dx}\:\Rightarrow\mathrm{I}\:=\int\:\frac{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)\left(\sqrt{\mathrm{2x}+\mathrm{3}}−\sqrt{\mathrm{x}+\mathrm{1}}\right)}{\mathrm{x}+\mathrm{2}}\mathrm{dx}…

tan-1-sinx-1-sinx-dx-

Question Number 29043 by yesaditya22@gmail.com last updated on 03/Feb/18 $$\int\mathrm{tan}^{−} \left(\mathrm{1}−\mathrm{sinx}/\mathrm{1}+\mathrm{sinx}\right)\:\mathrm{dx} \\ $$ Commented by abdo imad last updated on 03/Feb/18 $${let}\:{put}\:{I}=\:\int\:{arctan}\left(\frac{\mathrm{1}−{sinx}}{\mathrm{1}+{sinx}}\right){dx}\:\:\:\left({arctan}={tan}^{−\mathrm{1}} \right){we}\:{have} \\ $$$$\frac{\mathrm{1}−{sinx}}{\mathrm{1}+{sinx}}=\frac{\mathrm{1}\:−{cos}\left(\frac{\pi}{\mathrm{2}}−{x}\right)}{\mathrm{1}+{cos}\left(\frac{\pi}{\mathrm{2}}−{x}\right)}=\frac{\mathrm{2}{sin}^{\mathrm{2}}…