Question Number 200130 by universe last updated on 14/Nov/23 $$\:\:{solve}\:{by}\:{contour}\:{integrstion} \\ $$$$\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{dx}}{\mathrm{1}+{a}\mathrm{cos}{x}}\: \\ $$ Answered by Mathspace last updated on 14/Nov/23 $${I}=\int_{\mathrm{0}} ^{\mathrm{2}\pi}…
Question Number 200061 by universe last updated on 13/Nov/23 $$\:\:\:\:\int_{−\infty} ^{+\infty} \frac{{x}\mathrm{sin}{x}\:}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)}{dx}\:\:=\:\:\:?? \\ $$ Answered by witcher3 last updated on 13/Nov/23 $$\int_{−\infty} ^{\infty}…
Question Number 200048 by Calculusboy last updated on 12/Nov/23 Commented by 0670322918 last updated on 13/Nov/23 $$\int\frac{{tan}^{−\mathrm{1}} \left({x}\right)}{\int{tan}^{−\mathrm{1}} \left({x}\right){dx}}{dx}= \\ $$$${f}\left({x}\right)=\int{tan}^{−\mathrm{1}} \left({x}\right){dx}={xtan}^{−\mathrm{1}} \left({x}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}^{\mathrm{2}} +\mathrm{1}\right)+{c} \\…
Question Number 199934 by cortano12 last updated on 11/Nov/23 $$\:\:\:\cancel{\underline{\underbrace{ }}\:} \\ $$$$ \\ $$ Answered by mr W last updated on 11/Nov/23 $$\int_{\mathrm{0}} ^{\mathrm{3}}…
Question Number 199921 by Mingma last updated on 11/Nov/23 Answered by des_ last updated on 12/Nov/23 $$\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{cos}\:{x}}{{x}^{\mathrm{2}} +\mathrm{1}}\:{dx}\:=\:{I}; \\ $$$${I}\left({a}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{cos}\left({ax}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}\:{dx},\:{a}\:>\mathrm{0};…
Question Number 199907 by cortano12 last updated on 11/Nov/23 $$\:\:\:\:\:\int\:\frac{\mathrm{dx}}{\:\sqrt{\mathrm{x}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{x}}}\:=?\: \\ $$ Answered by Frix last updated on 11/Nov/23 $${t}=\sqrt[{\mathrm{6}}]{{x}} \\ $$$${x}={t}^{\mathrm{6}} \:\:\:{dx}=\mathrm{6}{x}^{\frac{\mathrm{5}}{\mathrm{6}}} {dt} \\…
Question Number 199903 by frankpenredpen last updated on 11/Nov/23 $$\int\frac{{x}^{\mathrm{2}} {dx}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{16}}}\:=\:? \\ $$ Commented by cortano12 last updated on 11/Nov/23 $$\mathrm{why}\:\partial\mathrm{x}\:? \\ $$ Answered…
Question Number 199942 by cortano12 last updated on 11/Nov/23 Answered by MM42 last updated on 11/Nov/23 $$\mathrm{1}−\sqrt{{x}}={u}^{\mathrm{2}} \Rightarrow{x}=\left(\mathrm{1}−{u}^{\mathrm{2}} \right)^{\mathrm{2}} ={x}\Rightarrow{dx}=−\mathrm{4}{u}\left(\mathrm{1}−{u}^{\mathrm{2}} \right){du} \\ $$$$\left.\Rightarrow{I}=\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \left({u}^{\mathrm{2}}…
Question Number 199570 by universe last updated on 05/Nov/23 $$\:\:\:\:\:\:\:\:\mathrm{I}\:\:\:\:\:=\:\:\:\:\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{sin}{x}\:}{\mathrm{2}}\right){dx} \\ $$ Answered by witcher3 last updated on 05/Nov/23 $$\mathrm{I}\left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{tan}^{−\mathrm{1}}…
Question Number 199598 by emilagazade last updated on 05/Nov/23 Terms of Service Privacy Policy Contact: info@tinkutara.com