Question Number 221350 by MrGaster last updated on 31/May/25 $$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{cos}\:\pi{x}}{\Gamma\left(\mathrm{2}+{x}\right)\Gamma\left(\mathrm{2}−{x}\right)}{dx} \\ $$$$ \\ $$$$ \\ $$ Commented by Ghisom last updated on 31/May/25…
Question Number 221256 by Nicholas666 last updated on 28/May/25 $$ \\ $$$$\:\:\int\int\int_{\:\left[\mathrm{0},\mathrm{1}\right]} \:\frac{\mathrm{ln}\left[\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+{y}\right)\left(\mathrm{1}+{z}\right]\right.}{\mathrm{1}\:+\:{xyz}}\:{dxdydz}\:\:\:\: \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 221254 by Nicholas666 last updated on 28/May/25 $$\:\:\int\int\int_{\left[\mathrm{0},\mathrm{1}\right]^{\mathrm{3}} } \:\frac{\mathrm{1}}{\left(\mathrm{1}−{xyz}\right)\left(\mathrm{1}−{xy}\right)\left(\mathrm{1}−{yz}\right)\left(\mathrm{1}−{zx}\right)}\:{dxdydz}\:\:\:\: \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 221203 by Nicholas666 last updated on 27/May/25 $$ \\ $$$$\:\:\:\:\:\int\:\mathrm{tan}\left(\frac{\frac{\mathrm{1}}{{x}}}{\mathrm{sec}\left({x}\right)}\right)\:+\:\frac{\mathrm{1}\:−\:\mathrm{sec}\left({x}\right)}{\mathrm{sec}\left({x}\right)}\:\mathrm{d}{x} \\ $$$$ \\ $$ Answered by SdC355 last updated on 27/May/25 $$\int\:\:\left[\mathrm{tan}\left(\frac{\mathrm{cos}\left({z}\right)}{{z}}\right)+\mathrm{cos}\left({z}\right)−\mathrm{1}\right]\:\mathrm{d}{z}=?? \\…
Question Number 221100 by Nicholas666 last updated on 24/May/25 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{Prove}}; \\ $$$$\:\:\int_{\mathrm{0}} ^{\:+\infty} \:\frac{\mathrm{4}\centerdot\boldsymbol{\mathrm{cos}}\:\boldsymbol{{x}}\:\centerdot\:\sqrt[{\mathrm{6}\:\:}]{\boldsymbol{\mathrm{sinh}}\:\boldsymbol{{x}}\:}}{\boldsymbol{\mathrm{sinh}}\:\boldsymbol{{x}}\:+\:\boldsymbol{\mathrm{sinh}}\:\mathrm{3}\boldsymbol{{x}}\:+\:\mathrm{4}\:\boldsymbol{\mathrm{sinh}}^{\mathrm{2}} \:\boldsymbol{{x}}\:−\:\mathrm{2}\:\boldsymbol{\mathrm{sinh}}^{\mathrm{2}} \:\mathrm{2}\boldsymbol{{x}}\:+\:\mathrm{4}\:\boldsymbol{\mathrm{sinh}}^{\mathrm{4}} \:\boldsymbol{{x}}\:+\:\mathrm{4}\:\boldsymbol{\mathrm{cosh}}^{\mathrm{4}} \boldsymbol{{x}}}\:\boldsymbol{\mathrm{d}{x}}\:=\:\frac{\boldsymbol{\pi}}{\:\sqrt{\mathrm{6}}\:+\:\mathrm{2}}\:\:\:\:\:\:\: \\ $$$$ \\ $$ Terms…
Question Number 221099 by Nicholas666 last updated on 24/May/25 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Prove}: \\ $$$$\:\:\underset{\:−\pi} {\overset{\:\pi} {\int}}\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{cos}^{{n}\:+\:\mathrm{1}} \:{x}}{\left({n}\:+\:\mathrm{1}\right)\left(\mathrm{1}\:+\:{e}^{{x}^{\mathrm{2}{n}\:+\mathrm{1}} } \right)}\:\:\mathrm{d}{x}\:=\:\pi\:\mathrm{ln2}\:\:\:\:\: \\ $$$$ \\ $$…
Question Number 221089 by shunmisaki007 last updated on 24/May/25 $$\mathrm{Find}\: \\ $$$$\left(\mathrm{1}\right)\:\int\frac{\mathrm{1}}{{x}^{\mathrm{9}} }{dx} \\ $$$$\left(\mathrm{2}\right)\:\int\frac{\mathrm{1}}{{x}^{\mathrm{9}} +\mathrm{1}}{dx}. \\ $$ Answered by MrGaster last updated on 24/May/25…
Question Number 221049 by Nicholas666 last updated on 23/May/25 $$ \\ $$$$\:\:\:\:\int\:\:\frac{\mathrm{cos}\:{x}}{\left({x}\:+\:\mathrm{cos}\:{x}\right)^{\mathrm{2}} }\:+\:\frac{\mathrm{2}\left(\mathrm{1}\:−\:\mathrm{sin}\:{x}\right)^{\mathrm{2}} }{\left({x}\:+\:\mathrm{cos}\:{x}\right)^{\mathrm{3}} }\:\:\mathrm{d}{x}\: \\ $$$$\: \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 221048 by fantastic last updated on 23/May/25 $$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\mathrm{cosec}\:\left({x}−\frac{\pi}{\mathrm{3}}\right)\mathrm{cosec}\:\left({x}−\frac{\pi}{\mathrm{6}}\right){dx}\: \\ $$ Answered by vnm last updated on 24/May/25 $$ \\ $$$$\mathrm{the}\:\mathrm{integral}\:\mathrm{diverges},\:\mathrm{but}\:\mathrm{it}'\mathrm{s} \\…
Question Number 221057 by fantastic last updated on 23/May/25 Commented by fantastic last updated on 23/May/25 $${Please}\:{find}\:{the}\:{area}\:{of}\:{shaded}\:{region}\:{using}\:{calculus} \\ $$ Answered by mr W last updated…