Question Number 65841 by Rio Michael last updated on 04/Aug/19 $$\:\frac{{d}}{{dx}}\left(\frac{{tan}\:^{\mathrm{2}} {x}}{\mathrm{1}\:+\:{cos}\:{x}}\right)\:=? \\ $$ Commented by som(math1967) last updated on 05/Aug/19 $${MJS}\:{Sir}\:{it}\:{is}\:{my}\:{way} \\ $$$$\frac{{d}}{{dx}}\left(\frac{{sec}^{\mathrm{2}} {x}−\mathrm{1}}{\mathrm{1}+{cosx}}\right)=\frac{{d}}{{dx}}\left\{\frac{\mathrm{1}−{cos}^{\mathrm{2}}…
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Question Number 65782 by Rio Michael last updated on 03/Aug/19 $$\:{Evaluate}\:\int_{\mathrm{0}} ^{\mathrm{2}} \left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}{x}\:+\:\mathrm{4}\right)^{\mathrm{7}} {dx} \\ $$$${hence}\:{show}\:{that}\:\:\frac{{d}}{{dx}}\left({coshx}\right)\:=\:{sinh}\:{x} \\ $$ Commented by mathmax by abdo last…
Question Number 204 by 02@>@0 last updated on 25/Jan/15 $${x}^{\mathrm{2}} +\left({y}−\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} }\right)^{\mathrm{2}} =\mathrm{1} \\ $$ Commented by 02@>@0 last updated on 15/Dec/14 $$ \\ $$…
Question Number 65735 by Adisco last updated on 03/Aug/19 $${show}\:{that}\:{the}\:{maping}\:{define}\:{by} \\ $$$$\left({x}\:{y}\right)=\Sigma{x}\bar {{y}} \\ $$$${is}\:{an}\:{inner}\:{product} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 65601 by AnjanDey last updated on 31/Jul/19 $$\mathrm{1}.\mathrm{If}\:\boldsymbol{{y}}=\boldsymbol{{x}}^{\boldsymbol{{n}}−\mathrm{1}} \mathrm{log}\:\boldsymbol{{x}},\mathrm{then}\:\mathrm{prove}\:\mathrm{that},\boldsymbol{{x}}^{\mathrm{2}} \frac{\boldsymbol{{d}}^{\mathrm{2}} \boldsymbol{{y}}}{\boldsymbol{{dx}}^{\mathrm{2}} }+\left(\mathrm{3}−\mathrm{2}\boldsymbol{{n}}\right)\boldsymbol{{x}}\frac{\boldsymbol{{dy}}}{\boldsymbol{{dx}}}+\left(\boldsymbol{{n}}−\mathrm{1}\right)^{\mathrm{2}} \boldsymbol{{y}}=\mathrm{0} \\ $$$$\mathrm{2}.\boldsymbol{\mathrm{I}}\mathrm{f}\:\frac{\mathrm{mtan}\:\left(\alpha−\theta\right)}{\mathrm{cos}\:^{\mathrm{2}} \theta}=\frac{{n}\mathrm{tan}\:\theta}{\mathrm{cos}\:^{\mathrm{2}} \left(\alpha−\theta\right)},\mathrm{then}\:\mathrm{prove}\:\mathrm{that},\theta=\frac{\mathrm{1}}{\mathrm{2}}\left[\alpha−\mathrm{tan}^{−\mathrm{1}} \left(\frac{{n}−{m}}{{n}+{m}}\mathrm{tan}\:\alpha\right)\right] \\ $$ Commented by Prithwish…
Question Number 65593 by Rio Michael last updated on 31/Jul/19 $$\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{3}}} \left(\mathrm{3}{x}\:+\:\mathrm{1}\right)^{\mathrm{5}} {dx}\:= \\ $$ Commented by mathmax by abdo last updated on 31/Jul/19…
Question Number 65592 by Rio Michael last updated on 31/Jul/19 $$\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{3}}} \left(\mathrm{3}{x}\:+\:\mathrm{1}\right)^{\mathrm{5}} {dx}\:= \\ $$ Answered by mr W last updated on 31/Jul/19 $$\int_{\mathrm{0}}…
Question Number 131094 by Chhing last updated on 01/Feb/21 $$\:\: \\ $$$$\:\:\:\mathrm{Calculate} \\ $$$$\:\:\mathrm{1}/\:\mathrm{I}\:=\:\oint_{\mathrm{c}^{+} } \frac{\mathrm{zdz}}{\left(\mathrm{z}−\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{z}^{\mathrm{2}} −\mathrm{2z}+\mathrm{1}−\mathrm{2i}\right)}\:\:,\mathrm{C}=\left\{\mathrm{z}/\mid\mathrm{z}\mid=\mathrm{2}\right\}\: \\ $$$$\:\:\mathrm{2}/\:\mathrm{J}\:=\oint_{\mathrm{c}^{+} } \frac{\mathrm{ch}\left(\mathrm{z}\right)\mathrm{dz}}{\mathrm{z}\left(\mathrm{e}^{\mathrm{z}} −\mathrm{1}\right)}\:\:,\:\:\mathrm{C}=\left\{\mathrm{z}/\mid\mathrm{z}−\mathrm{3i}\mid=\mathrm{4}\right\} \\ $$$$\:\:\mathrm{3}/\:\mathrm{K}=\oint_{\mathrm{c}^{+}…
Question Number 131083 by Dwaipayan Shikari last updated on 01/Feb/21 $$\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}−\frac{\mathrm{16}}{\mathrm{13}−\frac{\mathrm{81}}{\mathrm{25}−\frac{\mathrm{256}}{\mathrm{41}−\frac{\mathrm{625}}{\mathrm{61}−\frac{\mathrm{1296}}{\mathrm{85}−\frac{\mathrm{2401}}{\mathrm{113}−…}}}}}}}=\frac{\mathrm{6}}{\boldsymbol{\pi}^{\mathrm{2}} } \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com