Question Number 140487 by Dwaipayan Shikari last updated on 08/May/21 $$\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\right)\frac{{dx}}{{x}}={log}\left(\mathrm{2}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 74948 by vishalbhardwaj last updated on 04/Dec/19 $$\mathrm{The}\:\mathrm{hhpotenuse}\:\mathrm{of}\:\mathrm{a}\:\mathrm{right}\:\mathrm{angled}\:\mathrm{triangle} \\ $$$$\mathrm{has}\:\mathrm{its}\:\mathrm{ends}\:\mathrm{at}\:\mathrm{the}\:\mathrm{points}\:\left(\mathrm{1},\mathrm{3}\right)\:\mathrm{and}\:\left(−\mathrm{4},\mathrm{1}\right) \\ $$$$.\:\mathrm{Find}\:\mathrm{an}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{legs}\:\left(\mathrm{perpendicar}\right. \\ $$$$\left.\:\mathrm{sides}\right)\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangle}. \\ $$ Answered by MJS last updated on 04/Dec/19…
Question Number 140481 by ajfour last updated on 08/May/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 74947 by chess1 last updated on 04/Dec/19 Answered by MJS last updated on 05/Dec/19 $$\left({x}−\mathrm{1}\right)\left(\mathrm{8}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{1}\right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\Rightarrow\:{x}=\pm\mathrm{1}\vee{x}=−\frac{\mathrm{1}}{\mathrm{6}}−\frac{\sqrt{\mathrm{7}}}{\mathrm{3}}\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arcsin}\:\frac{\sqrt{\mathrm{7}}}{\mathrm{14}}\right)\:\vee{x}=−\frac{\mathrm{1}}{\mathrm{6}}+\frac{\sqrt{\mathrm{7}}}{\mathrm{3}}\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arcsin}\:\frac{\sqrt{\mathrm{7}}}{\mathrm{14}}\right)\:\vee{x}=−\frac{\mathrm{1}}{\mathrm{6}}−\frac{\sqrt{\mathrm{7}}}{\mathrm{3}}\mathrm{cos}\:\left(\frac{\pi}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arcsin}\:\frac{\sqrt{\mathrm{7}}}{\mathrm{14}}\right) \\ $$ Commented…
Question Number 74944 by aliesam last updated on 04/Dec/19 $$\int\frac{{e}^{−{cos}\left(\mathrm{2}{x}\right)} }{{sin}^{\mathrm{2}} \left({x}\right)}\:{dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 74945 by chess1 last updated on 04/Dec/19 Commented by chess1 last updated on 04/Dec/19 $$\mathrm{solve}\:\mathrm{equation} \\ $$ Answered by MJS last updated on…
Question Number 140477 by ajfour last updated on 08/May/21 $${If}\:\:\mathrm{sin}\:\theta+\mathrm{sin}\:^{\mathrm{2}} \theta+\mathrm{sin}\:^{\mathrm{3}} \theta+…. \\ $$$$\:\:\:\:\:\:=\:\mathrm{cos}\:\theta\:\:\:\:\:{and}\:\:\mathrm{0}<\theta<\frac{\pi}{\mathrm{2}}\:{then} \\ $$$${find}\:\theta. \\ $$ Answered by benjo_mathlover last updated on 08/May/21…
Question Number 9407 by geovane10math last updated on 06/Dec/16 $$\pi\:=\:\mathrm{180}°\:=\:\Pi \\ $$$$\pi\:=\:\mathrm{3},\mathrm{141}…\:=\:\pi \\ $$$$\mathrm{Solve}\:\left(\mathrm{find}\:{n}\:\in\:\mathbb{N}\right): \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{sin}\left(\mathrm{2}\Pi{k}\centerdot\mathrm{10}^{{n}} \pi\right)}{{k}}\:+\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{sin}\left(\mathrm{2}\Pi{k}\centerdot\mathrm{10}^{{n}} {e}\right)}{{k}}\:=\:\mathrm{0} \\ $$$$ \\…
Question Number 9404 by alfat123 last updated on 05/Dec/16 Answered by geovane10math last updated on 06/Dec/16 $$\mathrm{I}\:\mathrm{can}'\mathrm{t}\:\mathrm{read}… \\ $$$${f}'\left({x}\right)\:=\:\mathrm{2}{x}\:+\:\mathrm{2} \\ $$$$\int{f}'\left({x}\right)\:{dx}\:=\:\int\:\mathrm{2}{x}\:+\:\mathrm{2}\:{dx}\:=\:\int\mathrm{2}{x}\:{dx}\:+\:\int\mathrm{2}\:{dx} \\ $$$$=\:{x}^{\mathrm{2}} \:+\:\mathrm{2}{x}\:+\:{C} \\…
Question Number 9403 by alfat123 last updated on 05/Dec/16 Answered by ridwan balatif last updated on 05/Dec/16 $$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{5x}\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{6}} \mathrm{dx}=\mathrm{u}.\mathrm{v}−\int\mathrm{vdu} \\ $$$$\mathrm{misal}:\:\mathrm{u}=\mathrm{5x}\rightarrow\mathrm{du}=\mathrm{5dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{dv}=\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{6}}…