Question Number 126997 by bramlexs22 last updated on 26/Dec/20 $$\:\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\mathrm{arcsin}\:\left(\frac{\mathrm{sin}\:{x}}{\:\sqrt{\mathrm{2}}}\right)\:{dx}\:=? \\ $$ Answered by Evimene last updated on 26/Dec/20 $$\mathrm{solution} \\ $$$$\mathrm{let}\:\sqrt{\mathrm{2}}=\alpha \\…
Question Number 126994 by bramlexs22 last updated on 26/Dec/20 $$\:\int\:\frac{{x}^{\mathrm{4}} }{\mathrm{1}+{x}^{\mathrm{8}} }\:{dx}\:=? \\ $$ Answered by liberty last updated on 26/Dec/20 $$\Omega=\int\frac{{x}^{\mathrm{4}} }{\mathrm{1}+{x}^{\mathrm{8}} }\:{dx}\:=\:\int\frac{{dx}}{{x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}}…
Question Number 126992 by Tinku Tara last updated on 26/Dec/20 $$\mathrm{App}\:\mathrm{Information}: \\ $$$$\mathrm{We}\:\mathrm{have}\:\mathrm{unpublished}\:\mathrm{free}\:\mathrm{versoon} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{app}\:\mathrm{from}\:\mathrm{playstore}. \\ $$$$\mathrm{Existing}\:\mathrm{users}\:\mathrm{can}\:\mathrm{still}\:\mathrm{see}\:\mathrm{the} \\ $$$$\mathrm{app}\:\mathrm{on}\:\mathrm{playstore}. \\ $$$$\mathrm{A}\:\mathrm{paid}\:\mathrm{version}\:\mathrm{will}\:\mathrm{soon}\:\mathrm{be}\:\mathrm{available}. \\ $$$$\mathrm{We}\:\mathrm{will}\:\mathrm{update}\:\mathrm{pinned}\:\mathrm{message}\:\mathrm{once} \\ $$$$\mathrm{it}\:\mathrm{is}\:\mathrm{available}.…
Question Number 126990 by physicstutes last updated on 26/Dec/20 $$\mathrm{Obtain}\:\mathrm{a}\:\mathrm{formula}\:\mathrm{for}\: \\ $$$$\:{I}_{{n}} \:=\:\underset{\mathrm{0}} {\overset{{n}} {\int}}\left[{x}\right]\:{dx}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:{n} \\ $$$$\:\mathrm{where}\:\left[{x}\right]\:\mathrm{is}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{integer}\:\mathrm{function}\:\mathrm{of}\:{x}. \\ $$ Answered by mr W last updated…
Question Number 61453 by maxmathsup by imad last updated on 02/Jun/19 $${find}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left({x}\right){ln}\left(\mathrm{1}+{x}\right)}{{x}}{dx} \\ $$ Answered by Smail last updated on 02/Jun/19 $${A}=\int_{\mathrm{0}} ^{\mathrm{1}}…
Question Number 126989 by arash sharifi last updated on 25/Dec/20 $$\int{e}^{\sqrt{{x}}} {dx} \\ $$ Answered by bramlexs22 last updated on 26/Dec/20 $${let}\:\sqrt{{x}}\:=\:{u}\:\Rightarrow{x}\:=\:{u}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:{dx}\:=\:\mathrm{2}{u}\:{du}\: \\…
Question Number 126986 by mnjuly1970 last updated on 25/Dec/20 $$\:\:\:\:\:\:\:\:\:\:\:…\:{nice}\:\:{calculus}… \\ $$$$\:\:\:\:\:{prove}\:\:{that}\::: \\ $$$$\:\:\:\:\mathrm{I}\::=\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \frac{\left\{{cot}\left({x}\right)\right\}}{{cot}\left({x}\right)}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\left(\pi−{ln}\left(\frac{{sinh}\left(\pi\right)}{\pi}\right)\right) \\ $$$$\left\{{x}\right\}\:{is}\:{fractional}\:{part}\:{of}\:\:{x}\:.. \\ $$ Answered by Olaf last updated…
Question Number 61451 by Tawa1 last updated on 02/Jun/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 126987 by sdfg last updated on 25/Dec/20 Answered by mr W last updated on 26/Dec/20 $${Q}\mathrm{4} \\ $$$$\boldsymbol{{C}}=\alpha\boldsymbol{{A}}+\beta\boldsymbol{{B}} \\ $$$$\mid\boldsymbol{{C}}\mid=\sqrt{\alpha^{\mathrm{2}} \mid\boldsymbol{{A}}\mid^{\mathrm{2}} +\beta^{\mathrm{2}} \mid\boldsymbol{{B}}\mid^{\mathrm{2}}…
Question Number 61449 by Tawa1 last updated on 02/Jun/19 Commented by Tawa1 last updated on 02/Jun/19 $$\frac{\mathrm{a}^{\mathrm{2}} \:+\:\mathrm{b}^{\mathrm{2}} }{\mathrm{c}^{\mathrm{2}} }\:\:=\:\:? \\ $$ Answered by perlman…