Question Number 33128 by prof Abdo imad last updated on 10/Apr/18 $${find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\:\mathrm{1}+{x}^{\mathrm{4}} \right)}\:. \\ $$ Commented by prof Abdo imad last updated…
Question Number 33125 by prof Abdo imad last updated on 10/Apr/18 $${let}\:{give}\:{u}_{{n}} =\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\frac{{cos}\left({nx}\right){dx}}{\mathrm{1}−\mathrm{2}\lambda{cosx}\:+\lambda^{\mathrm{2}} } \\ $$$$\left.\mathrm{1}\right)\:{prove}\:{that}\:\:\lambda\:{u}_{{n}+\mathrm{2}} \:−\left(\mathrm{1}+\lambda^{\mathrm{2}} \right){u}_{{n}+\mathrm{1}} \:+\lambda\:{u}_{{n}} =\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\:{ptove}\:{that}\:\Sigma\:{u}_{{n}} \:{is}\:{convergent}\:{and}\:{find}\:{its}\:{sum}…
Question Number 33120 by abdo imad last updated on 10/Apr/18 $${let}\:{give}\:\alpha>\mathrm{0}\:{find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{dx}}{\:\sqrt{\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+\alpha{x}\right)}}\:. \\ $$ Answered by MJS last updated on 11/Apr/18 $$\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+\alpha{x}\right)=−\alpha{x}^{\mathrm{2}} +\left(\alpha−\mathrm{1}\right){x}+\mathrm{1} \\…
Question Number 33119 by abdo imad last updated on 10/Apr/18 $${find}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{{n}} }{{e}^{{t}} \:−\mathrm{1}}\:{dt}\:{by}\:{using}\:\xi\left({x}\right)\:{for}\:{n}\:{integr} \\ $$$$\xi\left({x}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{{x}} }\:\:\:{with}\:{x}>\mathrm{1}\:. \\ $$ Commented by prof…
Question Number 164163 by Zaynal last updated on 15/Jan/22 $$\mathrm{Prove}\:\mathrm{the}; \\ $$$$\int_{−\infty} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{1}\:+\:\boldsymbol{{x}}^{\mathrm{2}} }\:\boldsymbol{{dx}}\:=\:\boldsymbol{\pi} \\ $$$$\:^{\left\{\mathrm{Z}.\mathrm{A}\right\}} \\ $$ Answered by mathmax by abdo last…
Question Number 164162 by Zaynal last updated on 15/Jan/22 $$\mathrm{Prove}\:\mathrm{the}; \\ $$$$\left(\boldsymbol{{tan}}\:\boldsymbol{\alpha}\:+\:\frac{\boldsymbol{{cos}}\:\boldsymbol{\alpha}}{\mathrm{1}\:+\:\boldsymbol{{sin}}\:\boldsymbol{\alpha}}\right)\:\boldsymbol{{sin}}\:\boldsymbol{\alpha}\:=\:\boldsymbol{\alpha} \\ $$$$\:^{\left[\mathrm{Z}.\mathrm{A}\right]} \\ $$ Commented by som(math1967) last updated on 15/Jan/22 $$\left\{\boldsymbol{{tan}\alpha}\:+\frac{\boldsymbol{{cos}\alpha}\left(\mathrm{1}−\boldsymbol{{sin}\alpha}\right)}{\mathrm{1}−\boldsymbol{{sin}}^{\mathrm{2}} \boldsymbol{\alpha}}\right\}×{sin}\alpha…
Question Number 98623 by hardylanes last updated on 15/Jun/20 $${evaluate}\: \\ $$$$\int_{\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}} ^{\mathrm{2}} \frac{\mathrm{1}}{{x}^{\mathrm{2}} \sqrt{\mathrm{4}+{x}^{\mathrm{2}} }}{dx}\:{using}\:{the}\:{substitution}\:{x}=\mathrm{2tan}\theta \\ $$$$ \\ $$ Answered by Kunal12588 last updated…
Question Number 33069 by abdo imad last updated on 09/Apr/18 $${by}\:\:{using}\:{residus}\:{theorem}\:{prove}\:{that} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}\:{dt}\:=\:\frac{\pi}{{sin}\left(\pi{a}\right)}\:{with}\:\:\mathrm{0}<{a}<\mathrm{1}\:. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 164129 by mnjuly1970 last updated on 14/Jan/22 $$ \\ $$$$\:\:\:{prove} \\ $$$$ \\ $$$$\:\:\:\:\:\boldsymbol{\phi}=\:\mathscr{R}{e}\:\left(\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\mathrm{Li}_{\:\mathrm{2}} \:\left(\:\frac{\mathrm{1}}{{x}}\:\right)\:\right){dx}\:=\:\zeta\:\left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:−−−{m}.{n}−−− \\ $$ Answered by…
Question Number 98594 by bemath last updated on 15/Jun/20 Answered by bobhans last updated on 15/Jun/20 $$\int\:\mathrm{e}^{\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}} \:\left(\mathrm{x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{1}\right)\:\mathrm{dx}\:=\:\int\:\mathrm{e}^{\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}} \:\mathrm{x}^{\mathrm{2}} \:\mathrm{dx}\:+\:\int\:\mathrm{2x}\:\mathrm{e}^{\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}} \:\mathrm{dx}\:+\:\int\:\mathrm{e}^{\frac{\mathrm{x}^{\mathrm{2}}…