Question Number 151599 by mathdanisur last updated on 22/Aug/21 $$\Omega\:=\underset{\:\mathrm{0}} {\overset{\:\infty} {\int}}\mathrm{cos}\left(\mathrm{x}^{\boldsymbol{\mathrm{n}}} \right)\:\mathrm{dx}\:=\:? \\ $$ Answered by Lordose last updated on 22/Aug/21 $$ \\ $$$$\Omega\:\overset{\mathrm{x}=\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{n}}}…
Question Number 20524 by NECx last updated on 27/Aug/17 $$\int{x}^{{x}} {dx} \\ $$ Answered by Joel577 last updated on 28/Aug/17 $${It}\:{can}'{t}\:{be}\:{integrated} \\ $$ Terms of…
Question Number 20523 by Tinkutara last updated on 27/Aug/17 $$\mathrm{Let}\:\mathrm{the}\:\mathrm{sum}\:\underset{{n}=\mathrm{1}} {\overset{\mathrm{9}} {\sum}}\frac{\mathrm{1}}{{n}\left({n}\:+\:\mathrm{1}\right)\left({n}\:+\:\mathrm{2}\right)}\:\mathrm{written} \\ $$$$\mathrm{in}\:\mathrm{its}\:\mathrm{lowest}\:\mathrm{terms}\:\mathrm{be}\:\frac{{p}}{{q}}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{value} \\ $$$$\mathrm{of}\:{q}\:−\:{p}. \\ $$ Answered by ajfour last updated on 27/Aug/17…
Question Number 151585 by peter frank last updated on 22/Aug/21 $$\int\frac{\mathrm{dx}}{\mathrm{x}\sqrt{\mathrm{a}^{\mathrm{n}} +\mathrm{x}^{\mathrm{n}} }} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 151587 by peter frank last updated on 22/Aug/21 $$\int\mathrm{sin}^{−\mathrm{1}} \sqrt{\frac{\mathrm{x}}{\mathrm{a}+\mathrm{x}}}\:\mathrm{dx} \\ $$ Answered by MJS_new last updated on 22/Aug/21 $${u}'=\mathrm{1}\:\rightarrow\:{u}={x} \\ $$$${v}=\mathrm{arcsin}\:\sqrt{\frac{{x}}{{x}+{a}}}\:\rightarrow\:{v}'=\frac{\sqrt{{a}}}{\mathrm{2}\left({x}+{a}\right)\sqrt{{x}}} \\…
Question Number 151586 by peter frank last updated on 22/Aug/21 $$\int\mathrm{e}^{\mathrm{tan}^{−\mathrm{1}} \mathrm{x}} \left(\frac{\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\right)\mathrm{dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 20511 by dioph last updated on 27/Aug/17 $$\mathrm{Given}\:\mathrm{a}\:\mathrm{sphere}\:\mathrm{of}\:\mathrm{unit}\:\mathrm{radius}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{expression}\:\mathrm{of}\:\mathrm{a}\:\mathrm{circular} \\ $$$$\mathrm{spot}\:\mathrm{on}\:\mathrm{the}\:\mathrm{sphere}'\mathrm{s}\:\mathrm{surface}\:\mathrm{given} \\ $$$$\mathrm{the}\:\mathrm{latitude}\:\beta\:\mathrm{and}\:\mathrm{the}\:\mathrm{longitude}\:\lambda \\ $$$$\mathrm{of}\:\mathrm{its}\:\mathrm{center}\:\mathrm{and}\:\mathrm{its}\:\mathrm{angular}\:\mathrm{radius}\:{r}. \\ $$ Terms of Service Privacy Policy…
Question Number 20509 by Tinkutara last updated on 27/Aug/17 $${Simplify}: \\ $$$$\mathrm{cos}^{−\mathrm{1}} \:\left(\frac{\mathrm{sin}\:{x}\:+\:\mathrm{cos}\:{x}}{\:\sqrt{\mathrm{2}}}\right),\:\frac{\mathrm{5}\pi}{\mathrm{4}}\:<\:{x}\:<\:\frac{\mathrm{9}\pi}{\mathrm{4}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 86042 by M±th+et£s last updated on 26/Mar/20 $${solve}:\:\:\lfloor\:\sqrt{{x}}\:\rfloor=\lfloor\frac{{x}}{\mathrm{2}}\rfloor \\ $$ Answered by Rio Michael last updated on 26/Mar/20 $$\:\lfloor\sqrt{{x}}\:\rfloor\:=\:\lfloor\frac{{x}}{\mathrm{2}}\rfloor \\ $$$$\:\Rightarrow\:\sqrt{{x}}\:=\:\frac{{x}}{\mathrm{2}} \\ $$$$\:\:\:{x}\:=\:\frac{{x}^{\mathrm{2}}…
Question Number 20506 by Tinkutara last updated on 27/Aug/17 $${Simplify}: \\ $$$$\mathrm{cos}^{−\mathrm{1}} \:\left(\frac{\mathrm{sin}\:{x}\:+\:\mathrm{cos}\:{x}}{\:\sqrt{\mathrm{2}}}\right),\:\frac{\pi}{\mathrm{4}}\:<\:{x}\:<\:\frac{\mathrm{5}\pi}{\mathrm{4}} \\ $$ Answered by ajfour last updated on 27/Aug/17 $$\:\theta=\mathrm{cos}^{−\mathrm{1}} \left[\mathrm{cos}\:\left(\pi/\mathrm{4}\right)\mathrm{cos}\:{x}+\mathrm{sin}\:\left(\pi/\mathrm{4}\right)\mathrm{sin}\:{x}\right] \\…