Question Number 62907 by aliesam last updated on 26/Jun/19 $$\int\sqrt{\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\:}\left(\mathrm{1}+{x}\right)\:{dx} \\ $$ Commented by mathmax by abdo last updated on 26/Jun/19 $${let}\:{A}\:=\int\sqrt{\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}}\left(\mathrm{1}+{x}\right){dx}\:{changement}\:{x}\:={cos}\left(\mathrm{2}\theta\right)\:{give}\:\mathrm{2}\theta\:={arcosx}\:\Rightarrow \\ $$$${A}\:=\int\:\sqrt{\frac{\mathrm{2}{cos}^{\mathrm{2}} \left(\theta\right)}{\mathrm{2}{sin}^{\mathrm{2}}…
Question Number 62906 by Prithwish sen last updated on 26/Jun/19 $$\int\mathrm{x}^{\mathrm{x}} \mathrm{dx} \\ $$ Commented by mathmax by abdo last updated on 27/Jun/19 $${let}\:{A}\:=\int\:{x}^{{x}} \:{dx}\:\Rightarrow{A}\:=\int\:{e}^{{xln}\left({x}\right)}…
Question Number 128439 by Study last updated on 07/Jan/21 $${li}\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {{m}}\frac{{cosx}}{\mathrm{1}−{sinx}}=??? \\ $$ Answered by benjo_mathlover last updated on 07/Jan/21 $$\mathrm{let}\:\mathrm{x}=\frac{\pi}{\mathrm{2}}+\mathrm{z}\:\wedge\:\mathrm{z}\rightarrow\mathrm{0} \\ $$$$\:\underset{\mathrm{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}+\mathrm{z}\right)}{\mathrm{1}−\mathrm{sin}\:\left(\mathrm{z}+\frac{\pi}{\mathrm{2}}\right)}=\underset{\mathrm{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{sin}\:\mathrm{z}}{\mathrm{1}−\mathrm{cos}\:\mathrm{z}}…
Question Number 128432 by bemath last updated on 07/Jan/21 $$\left(\mathrm{3x}^{\mathrm{2}} +\mathrm{1}\right)\mathrm{yy}'\:+\mathrm{3xy}^{\mathrm{2}} =\mathrm{cos}\:\mathrm{x}.\mathrm{e}^{\mathrm{sin}\:\mathrm{x}} \: \\ $$ Answered by liberty last updated on 07/Jan/21 $$\mathrm{let}\:\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{v}\left(\mathrm{x}\right)\Rightarrow\mathrm{2yy}'=\frac{\mathrm{dv}}{\mathrm{dx}} \\…
Question Number 128430 by sachin1221 last updated on 07/Jan/21 $$\boldsymbol{{if}}\:\boldsymbol{{we}}\:\boldsymbol{{have}}\:\boldsymbol{{DE}} \\ $$$$\frac{\boldsymbol{{d}}^{\mathrm{2}} \boldsymbol{{x}}\left({t}\right)}{\boldsymbol{{dt}}^{\mathrm{2}} }\:=\:−\boldsymbol{{sinx}} \\ $$$$\boldsymbol{{then}}\:\boldsymbol{{find}}\:\:\boldsymbol{{x}}\left(\boldsymbol{{t}}\right) \\ $$$$\mathrm{1}.\:\boldsymbol{{x}}\left(\boldsymbol{{t}}\right)=\:\boldsymbol{{acosht}}+\boldsymbol{{bsinht}} \\ $$$$\mathrm{2}.\:\boldsymbol{{x}}\left(\boldsymbol{{t}}\right)=\:\boldsymbol{{a}}+\boldsymbol{{bt}} \\ $$$$\mathrm{3}.\:\boldsymbol{{x}}\left(\boldsymbol{{t}}\right)=\:\boldsymbol{{ae}}^{\boldsymbol{{t}}} \:+\:\boldsymbol{{be}}^{\mathrm{2}\boldsymbol{{t}}} \\ $$$$\mathrm{4}.\:\boldsymbol{{x}}\left(\boldsymbol{{t}}\right)=\:\boldsymbol{{acost}}\:+\:\boldsymbol{{bsint}}…
Question Number 62895 by Tawa1 last updated on 26/Jun/19 Commented by Tony Lin last updated on 26/Jun/19 $${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\Rightarrow{a}=\pm\sqrt{\frac{\mathrm{1}}{\mathrm{2}}},\:{b}=\pm\sqrt{\frac{\mathrm{1}}{\mathrm{2}}},\:{c}=\pm\sqrt{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$\because\:\mid{ab}\mid<\mid{bc}\mid=\mid{ac}\mid…
Question Number 128424 by bramlexs22 last updated on 07/Jan/21 $$\mathrm{If}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{an}\:\mathrm{even}\:\mathrm{function}\:\mathrm{and} \\ $$$$\mathrm{satisfies}\:\mathrm{the}\:\mathrm{relation}\:\mathrm{x}^{\mathrm{2}} \mathrm{f}\left(\mathrm{x}\right)−\mathrm{2f}\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{x}\right) \\ $$$$\mathrm{where}\:\mathrm{g}\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{an}\:\mathrm{odd}\:\mathrm{function}\: \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{5}\right)\:\mathrm{is}\: \\ $$$$\left(\mathrm{a}\right)\:\mathrm{0}\:\:\left(\mathrm{b}\right)\:\frac{\mathrm{37}}{\mathrm{75}}\:\:\:\left(\mathrm{c}\right)\:\mathrm{4}\:\:\:\:\left(\mathrm{d}\right)\:\frac{\mathrm{51}}{\mathrm{77}} \\ $$ Answered by mr W…
Question Number 62889 by Tawa1 last updated on 26/Jun/19 Answered by Kunal12588 last updated on 26/Jun/19 $$\left({a}\right){s}={t}^{\mathrm{2}} −\mathrm{5}{t}+\mathrm{6} \\ $$$${at}\:{initial}\:{pos}.\:{t}=\mathrm{0} \\ $$$${s}\left(\mathrm{0}\right)=\mathrm{6} \\ $$$$\therefore{the}\:{initial}\:{dist}.\:{of}\:{particle}\:{from}\:{O}\:{is}\:\mathrm{6}{m} \\…
Question Number 128425 by bramlexs22 last updated on 07/Jan/21 $$\:\mathrm{If}\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{x}\:+\:\mathrm{tan}\:\mathrm{x}\:\mathrm{and}\:\mathrm{f}\:\mathrm{is}\:\mathrm{inverse} \\ $$$$\mathrm{of}\:\mathrm{g}\:,\:\mathrm{then}\:\mathrm{g}'\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$$$\left(\mathrm{a}\right)\:\frac{\mathrm{1}}{\mathrm{1}+\left(\mathrm{g}\left(\mathrm{x}\right)−\mathrm{x}\right)^{\mathrm{2}} }\:\:\:\left(\mathrm{b}\right)\:\frac{\mathrm{1}}{\mathrm{1}−\left(\mathrm{g}\left(\mathrm{x}\right)−\mathrm{x}\right)^{\mathrm{2}} } \\ $$$$\left(\mathrm{c}\right)\:\frac{\mathrm{1}}{\mathrm{2}+\left(\mathrm{g}\left(\mathrm{x}\right)−\mathrm{x}\right)^{\mathrm{2}} }\:\:\:\left(\mathrm{d}\right)\:\frac{\mathrm{1}}{\mathrm{2}−\left(\mathrm{g}\left(\mathrm{x}\right)−\mathrm{x}\right)^{\mathrm{2}} } \\ $$ Answered by liberty…
Question Number 128422 by Khalmohmmad last updated on 07/Jan/21 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{{n}^{{a}} }{{b}^{{n}} }\:\:\:\:\:\:\:{a}>\mathrm{0}\:\:,\:\:{b}>\mathrm{1} \\ $$ Answered by mathmax by abdo last updated on 07/Jan/21 $$\mathrm{let}\:\mathrm{u}_{\mathrm{n}}…