Question Number 211902 by Gangadhar last updated on 23/Sep/24 Commented by BHOOPENDRA last updated on 24/Sep/24 $$\mathrm{3} \\ $$ Answered by BHOOPENDRA last updated on…
Question Number 211896 by Spillover last updated on 23/Sep/24 Answered by BHOOPENDRA last updated on 23/Sep/24 $$=−\frac{\mathrm{1}}{\mathrm{2}}\left({Re}\underset{{z}=\mathrm{0}} {{s}}\left({f}\left({z}\right)\right)+{Res}\underset{{z}=−\mathrm{1}} {\:}\:\left({f}\left({z}\right)\right)\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{n}−\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{2}{n}^{\mathrm{2}} +\mathrm{1}\right)\right) \\ $$$$=\frac{\mathrm{2}{n}^{\mathrm{2}} −\mathrm{3}{n}+\mathrm{1}}{\mathrm{3}}…
Question Number 211895 by Spillover last updated on 23/Sep/24 Answered by mehdee7396 last updated on 23/Sep/24 $${lim}_{{x}\rightarrow\infty} \:\left(\mathrm{1}+\frac{{a}}{{x}}−\frac{\mathrm{4}}{{x}^{\mathrm{2}} }−\mathrm{1}\right)\mathrm{2}{x} \\ $$$$={lim}_{{x}\rightarrow\infty} \:\left(\frac{{ax}−\mathrm{4}}{{x}^{\mathrm{2}} }\right)\mathrm{2}{x}=\mathrm{2}{a} \\ $$$$\Rightarrow{lim}_{{x}\rightarrow\infty}…
Question Number 211885 by Spillover last updated on 23/Sep/24 Answered by Frix last updated on 23/Sep/24 $$\int\frac{{x}^{\mathrm{2}} }{\left({x}\mathrm{sin}\:{x}\:+\mathrm{cos}\:{x}\right)^{\mathrm{2}} }{dx}=\frac{{y}}{{x}\mathrm{sin}\:{x}\:+\mathrm{cos}\:{x}}+{C} \\ $$$$\frac{{d}}{{dx}}\left[\frac{{y}}{{x}\mathrm{sin}\:{x}\:+\mathrm{cos}\:{x}}\right]=\frac{{x}^{\mathrm{2}} }{\left({x}\mathrm{sin}\:{x}\:+\mathrm{cos}\:{x}\right)^{\mathrm{2}} } \\ $$$$\frac{{y}'\left({x}\mathrm{sin}\:{x}\:+\mathrm{cos}\:{x}\right)−{yx}\mathrm{cos}\:{x}}{\left({x}\mathrm{sin}\:{x}\:+\mathrm{cos}\:{x}\right)^{\mathrm{2}}…
Question Number 211886 by Spillover last updated on 23/Sep/24 Answered by Ghisom last updated on 23/Sep/24 $$\int\sqrt{\mathrm{cos}^{\mathrm{2}} \:{x}\:+\mathrm{cos}\:{x}}\:{dx}=\int\sqrt{\mathrm{2cos}\:{x}}\:\mathrm{cos}\:\frac{{x}}{\mathrm{2}}\:{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{\sqrt{\mathrm{2}}\mathrm{sin}\:\frac{{x}}{\mathrm{2}}}{\:\sqrt{\mathrm{cos}\:{x}}}\:\Leftrightarrow\:{x}=\mathrm{2arctan}\:\frac{{t}}{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{2}}}\:\rightarrow\:{dx}=\frac{\sqrt{\mathrm{2cos}^{\mathrm{3}} \:{x}}}{\mathrm{cos}\:\frac{{x}}{\mathrm{2}}}{dt}\right] \\ $$$$=\mathrm{2}\int\frac{{dt}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}}…
Question Number 211880 by Spillover last updated on 23/Sep/24 Answered by Frix last updated on 23/Sep/24 $$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{\mathrm{tan}\:{x}}{\pi^{\mathrm{2}} +\mathrm{4ln}^{\mathrm{2}} \:\mathrm{tan}\:{x}}{dx}\:\overset{\left[{t}=\mathrm{2ln}\:\mathrm{tan}\:{x}\right]} {=} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{−\infty} {\overset{\infty}…
Question Number 211881 by Spillover last updated on 23/Sep/24 Answered by som(math1967) last updated on 23/Sep/24 $$\:\alpha+\beta+\gamma=\mathrm{1},\alpha\beta+\beta\gamma+\gamma\alpha=−\mathrm{2} \\ $$$$\:\alpha\beta\gamma=−\mathrm{1} \\ $$$$\:\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} +\gamma^{\mathrm{2}} =\mathrm{1}+\mathrm{4}=\mathrm{5} \\…
Question Number 211908 by Spillover last updated on 23/Sep/24 Answered by BHOOPENDRA last updated on 24/Sep/24 $${I}=\int_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} \left(\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{sin}\:\left(\frac{{k}\pi}{\mathrm{2}}\right){x}^{{k}} \right){dx} \\ $$$${The}\:{pattern}\:{for}\:\mathrm{sin}\:\left({k}\pi/\mathrm{2}\right) \\…
Question Number 211879 by Spillover last updated on 23/Sep/24 Answered by BHOOPENDRA last updated on 23/Sep/24 $$=\:\frac{\left({x}−\mathrm{4}\right)}{\mathrm{10}\left({x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{5}\right)}\:+\frac{\mathrm{1}}{\mathrm{10}}\int\:\frac{{x}−\mathrm{4}}{{x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} +\mathrm{5}{x}}\:\left({by}\:{ostrogradsky}'{s}\:{method}\right) \\ $$$$\:\:=\frac{{x}−\mathrm{4}}{\mathrm{10}\left({x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{5}\right)}+\frac{\mathrm{1}}{\mathrm{10}}\left[\int\frac{\mathrm{4}{x}−\mathrm{11}}{\mathrm{5}\left({x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{5}\right)}{dx}−\int\frac{\mathrm{4}}{\mathrm{5}{x}}\:{dx}\right] \\…
Question Number 211837 by Majak last updated on 22/Sep/24 Terms of Service Privacy Policy Contact: info@tinkutara.com