Question Number 140715 by mnjuly1970 last updated on 11/May/21 $$\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:……{advanced}\:\:{calculus}…… \\ $$$$\:\:\:\:\:\:{prove}\:{that}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}:=\int_{\mathrm{0}} ^{\:\infty} \frac{{x}^{\mathrm{2}} }{{cosh}^{\mathrm{2}} \left({x}^{\mathrm{2}} \right)}{dx}\overset{?} {=}\frac{\sqrt{\mathrm{2}}\:−\mathrm{2}}{\mathrm{4}}\:\sqrt{\pi}\:\zeta\:\left(\:\frac{\mathrm{1}}{\mathrm{2}}\:\right)\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:………….. \\…
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Question Number 75177 by chess1 last updated on 08/Dec/19 Commented by chess1 last updated on 08/Dec/19 $$\mathrm{sir}\:\mathrm{mind}\:\mathrm{is}\:\mathrm{power}\:\:\mathrm{please}\:\mathrm{solution} \\ $$ Commented by mind is power last…
Question Number 140703 by rs4089 last updated on 11/May/21 Answered by Dwaipayan Shikari last updated on 11/May/21 $${x}={t}+\mathrm{1} \\ $$$$\int_{\mathrm{1}} ^{\infty} \frac{{dt}}{\left({t}+\mathrm{1}\right)^{{p}+\mathrm{1}} {t}^{{q}} }\:\:\:{t}=\frac{\mathrm{1}}{{g}} \\…
Question Number 9623 by tawakalitu last updated on 21/Dec/16 $$\mathrm{Evaluate}\::\:\int_{\mathrm{4}} ^{\mathrm{5}.\mathrm{2}} \:\mathrm{ln}\left(\mathrm{x}\right)\:\mathrm{dx}\:\: \\ $$$$\mathrm{using}\:\mathrm{trapezoidal}\:\mathrm{rule}.\:\mathrm{take}\:\mathrm{h}\:=\:\mathrm{0}.\mathrm{2} \\ $$ Commented by sandy_suhendra last updated on 21/Dec/16 Answered by…
Question Number 140684 by qaz last updated on 11/May/21 $$\int_{\mathrm{0}} ^{\pi} \mathrm{cos}\:^{{n}} \left({x}\right)\centerdot\mathrm{cos}\:\left({nx}\right){dx}=\frac{\pi}{\mathrm{2}^{{n}} } \\ $$ Answered by mnjuly1970 last updated on 11/May/21 $$\:\boldsymbol{\phi}:={Re}\left(\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\:\mathrm{2}\pi}…
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Question Number 140687 by mnjuly1970 last updated on 11/May/21 $$\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\: \\ $$ Commented by qaz last updated…
Question Number 75141 by chess1 last updated on 07/Dec/19 Commented by chess1 last updated on 08/Dec/19 $$\mathrm{solve}\:\mathrm{please} \\ $$ Commented by chess1 last updated on…
Question Number 9581 by dc630248@gmail.com last updated on 18/Dec/16 $$\int\mathrm{sin}^{\mathrm{4}} {x}.\mathrm{cos}^{\mathrm{4}} {xdx} \\ $$ Answered by mrW last updated on 18/Dec/16 $$\mathrm{sin}^{\mathrm{4}} {x}.\mathrm{cos}^{\mathrm{4}} {x}=\frac{\mathrm{1}}{\mathrm{16}}\left(\mathrm{2sinxcosx}\right)^{\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{16}}\mathrm{sin}^{\mathrm{4}}…