Question Number 35616 by abdo mathsup 649 cc last updated on 21/May/18 $${integrate}\:{the}\:{d}.{e}\:\:{y}^{''} \:−\mathrm{2}{y}^{'} \:+{y}\:=\:{x}^{\mathrm{2}} {ch}\left({x}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 166684 by cortano1 last updated on 25/Feb/22 $$\:\:\:\:\:\:\mathrm{T}\:=\:\int\:\frac{\mathrm{sin}\:\left(\mathrm{x}^{\mathrm{2}} +\mathrm{2}\right)}{\mathrm{2x}+\mathrm{4}}\:\mathrm{dx}=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 35615 by abdo mathsup 649 cc last updated on 21/May/18 $${let}\:\:{S}_{{n}} \:=\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\:\frac{\mathrm{1}}{\mathrm{3}{k}+\mathrm{1}} \\ $$$${calculate}\:{S}_{{n}} \:\:\:{interms}\:{of}\:{H}_{{n}} \:\:\:{with}\:{H}_{{n}} \:=\sum_{{k}=\mathrm{1}} ^{{n}} \frac{\mathrm{1}}{{k}} \\ $$…
Question Number 35613 by abdo mathsup 649 cc last updated on 21/May/18 $${find}\:\:{I}_{{a},{b}} =\:\int_{−\infty} ^{+\infty} \:\:\:\:\:\:\frac{{e}^{{x}} }{\left(\mathrm{1}+{a}\:{e}^{{x}} \right)\left(\mathrm{1}+{be}^{{x}} \right)}{dx}\:.. \\ $$ Terms of Service Privacy…
Question Number 35611 by abdo mathsup 649 cc last updated on 21/May/18 $${let}\:{h}\left({t}\right)\:=\:{e}^{{t}−{e}^{{t}} } \:\:\:\:{and}\:{for}\:{n}\geqslant\mathrm{0}\:{we}\:{put} \\ $$$${h}_{{n}} \left({t}\right)\:={nh}\left({nt}\right) \\ $$$${calculate}\:\:\int_{−\infty} ^{+\infty} \:{h}_{{n}} \left({t}\right){dt}\:. \\ $$…
Question Number 35612 by abdo mathsup 649 cc last updated on 21/May/18 $${calculate}\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\left(\mathrm{1}+{t}\right)^{−\frac{\mathrm{1}}{\mathrm{4}}} \:\:\:−\left(\mathrm{1}+{t}\right)^{−\frac{\mathrm{3}}{\mathrm{4}}} }{{t}}{dt}\: \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 35610 by abdo mathsup 649 cc last updated on 21/May/18 $$\left.{let}\:{give}\:{x}\in\right]\mathrm{0},\mathrm{2}\pi\left[\:\:{and}\:{a}\:\in{R},{b}\in\:{R}\right. \\ $$$${prove}\:{that}\:\:\frac{\pi−{x}}{\mathrm{2}}\:=\:{arctan}\left(\frac{{sinx}}{\mathrm{1}−{cosx}}\right) \\ $$$$\left.\mathrm{2}\right)\:{prove}\:{that}\:\mid{arctan}\left({a}\right)−{arctan}\left({b}\right)\mid\leqslant\mid{a}−{b}\mid \\ $$$$\left.\mathrm{3}\left.\right){let}\theta\:\in\right]\mathrm{0},\frac{\pi}{\mathrm{2}}\left[\:\:,\:{x}\:\in\left[\theta,\mathrm{2}\pi−\theta\right]\:,\:{r}\in\left[\mathrm{0},\mathrm{1}\left[\:{prove}\:{that}\right.\right.\right. \\ $$$$\mid\varphi\left({x},{r}\right)\:−\frac{\pi−{x}}{\mathrm{2}}\mid\leqslant\:\:\frac{\mathrm{1}−{r}}{\left(\mathrm{1}−{cos}\theta\right)^{\mathrm{2}} } \\ $$ Terms…
Question Number 35608 by abdo mathsup 649 cc last updated on 21/May/18 $${let}\:{r}\in\left[\mathrm{0},\mathrm{1}\left[\:{and}\:{x}\:{from}\:{R}\right.\right. \\ $$$${F}\left({x},{r}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}\pi}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\frac{\left(\mathrm{1}−{r}^{\mathrm{2}} \right){f}\left({t}\right)}{\mathrm{1}−\mathrm{2}{r}\:{cos}\left({t}−{x}\right)\:+{r}^{\mathrm{2}} }{dt}\:\:{with} \\ $$$${f}\:\:\in\:{C}^{\mathrm{0}} \left({R}\right)\:\:\mathrm{2}\pi\:{periodic}\:\:{and}\:\:\mid\mid{f}\mid\mid={sup}_{{t}\in{R}} \mid{f}\left({t}\right)\mid \\ $$$$\:{prove}\:{that}\:{F}\left({x},{r}\right)=\:\frac{{a}_{\mathrm{0}}…
Question Number 35609 by abdo mathsup 649 cc last updated on 21/May/18 $${let}\:{r}\:\in\left[\mathrm{0},\mathrm{1}\left[\:{and}\:{x}\in\:{R}\:\:{and}\:\right.\right. \\ $$$$\varphi\left({x},{r}\right)\:=\:{arctan}\left(\:\frac{{rsinx}}{\mathrm{1}−{r}\:{cosx}}\right) \\ $$$$\left.\mathrm{1}\right)\:{prove}\:{that}\:\:\frac{\partial\varphi}{\partial{x}}\left({x},{r}\right)\:\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:{r}^{{n}} \:{cos}\left({nx}\right) \\ $$$$\left.\mathrm{2}\right){prove}\:{that}\:\varphi\left({x},{r}\right)\:=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:{r}^{{n}} \:\:\frac{{sin}\left({nx}\right)}{{n}}…
Question Number 35605 by abdo mathsup 649 cc last updated on 21/May/18 $${let}\:{r}\in\left[\mathrm{0},\mathrm{1}\left[\:{and}\:\theta\:\in\:{R},{x}\in\:{R}\:{prove}\:{that}\right.\right. \\ $$$$\left.\mathrm{1}\right)\:\mathrm{1}+\:\mathrm{2}\:\sum_{{n}=\mathrm{1}} ^{+\infty} \:{r}^{{n}} {cos}\theta\:=\:\frac{\mathrm{1}−{r}^{\mathrm{2}} }{\mathrm{1}−\mathrm{2}{r}\:{cos}\theta\:+{r}^{\mathrm{2}} } \\ $$$$\left.\mathrm{2}\right)\mathrm{1}\:=\frac{\mathrm{1}}{\mathrm{2}\pi}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\:\:\frac{\left(\mathrm{1}−{r}^{\mathrm{2}} \right)}{\mathrm{1}−\mathrm{2}{rcos}\left({t}−{x}\right)\:+{r}^{\mathrm{2}}…