Question Number 186910 by Spillover last updated on 11/Feb/23 $$ \\ $$$$\:\:\:\:\:\:\int_{\mathrm{0}} ^{{a}} \sqrt{\frac{\mathrm{cos}\:\mathrm{2}{x}−\mathrm{cos}\:\mathrm{2}{a}}{\mathrm{cos}\:\mathrm{2}{x}+\mathrm{1}}}\:{dx}=\frac{\pi}{\mathrm{2}}\left(\mathrm{1}−\mathrm{cos}\:{a}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$ Answered by witcher3…
Question Number 55834 by Tawa1 last updated on 04/Mar/19 Answered by ajfour last updated on 04/Mar/19 $$\:\:\:\left(\mathrm{2}−\mathrm{1}\right){f}_{{min}} \left({x}=\mathrm{2}\right)\leqslant\int_{\mathrm{1}} ^{\:\:\mathrm{2}} {f}\left({x}\right){dx}\:\leqslant\:\frac{\left(\mathrm{2}−\mathrm{1}\right)}{\mathrm{2}}\left[{f}\left(\mathrm{1}\right)+{f}\left(\mathrm{2}\right)\right] \\ $$$$\:\:\:\Rightarrow\:\frac{\mathrm{1}}{\mathrm{17}}\:\leqslant\int_{\mathrm{1}} ^{\:\:\mathrm{2}} \frac{{dx}}{\mathrm{1}+{x}^{\mathrm{4}} }\:\leqslant\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{17}}\right)\:<\:\frac{\mathrm{7}}{\mathrm{24}}\:.…
Question Number 186856 by horsebrand11 last updated on 11/Feb/23 $$\int\:\frac{\sqrt{{x}+\mathrm{3}}+\sqrt{{x}+\mathrm{2}}}{\:\sqrt{{x}+\mathrm{3}}+\sqrt{{x}−\mathrm{2}}}\:{dx}\:=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 121314 by Bird last updated on 06/Nov/20 $${find}\:{lim}_{{n}\rightarrow+\infty} \:\int_{\mathrm{0}} ^{{n}} \:\:\:\frac{{cos}\left({nx}\right)}{{x}^{\mathrm{2}} +{n}^{\mathrm{2}} }{dx} \\ $$ Answered by mathmax by abdo last updated on…
Question Number 121315 by Bird last updated on 06/Nov/20 $${find}\:{lim}\:\int_{\frac{\mathrm{1}}{{n}}} ^{{n}} {arctan}\left(\mathrm{1}+\frac{{x}}{{n}}\right){e}^{−{nx}} {dx} \\ $$ Answered by Lordose last updated on 06/Nov/20 $$\mathrm{u}\:=\:\left(\mathrm{1}+\frac{\mathrm{x}}{\mathrm{n}}\right)\:\Rightarrow\:\mathrm{du}\:=\:\frac{\mathrm{dx}}{\mathrm{n}} \\ $$$$\mathrm{x}\:=\:\mathrm{n}\left(\mathrm{u}−\mathrm{1}\right)…
Question Number 186840 by cortano12 last updated on 11/Feb/23 $$\:\:\underset{\mathrm{9}} {\overset{\:\mathrm{16}} {\int}}\:\frac{\sqrt{\mathrm{4}−\sqrt{{x}}}}{{x}}\:{dx}\:=? \\ $$ Answered by horsebrand11 last updated on 11/Feb/23 $$\:\:{I}=\underset{\mathrm{9}} {\overset{\:\mathrm{16}} {\int}}\:\frac{\sqrt{\mathrm{4}−\sqrt{{x}}}}{{x}}\:{dx}\:=? \\…
Question Number 121300 by liberty last updated on 06/Nov/20 $$\:\:\int\:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}\:? \\ $$ Commented by liberty last updated on 06/Nov/20 Answered by TANMAY PANACEA…
Question Number 55759 by maxmathsup by imad last updated on 03/Mar/19 $${calculate}\:{I}\:=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{cost}}{\mathrm{3}\:+{sin}\left(\mathrm{2}{t}\right)}{dt}\:{and}\:{J}\:=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{sint}}{\mathrm{3}\:+{cos}\left(\mathrm{2}{t}\right)}{dt}\:. \\ $$ Answered by MJS last updated on 04/Mar/19…
Question Number 55760 by maxmathsup by imad last updated on 03/Mar/19 $${let}\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{cos}\left({xt}\right)}{\left({xt}^{\mathrm{2}} +{i}\right)^{\mathrm{2}} }{dx}\:\:\:{with}\:{x}\:{from}\:{R}\:\:{and}\:{x}\neq\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{explicit}\:{form}\:{of}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{extract}\:\:{A}\:={Re}\left({f}\left({x}\right)\right)\:{and}\:\:{B}\:={Im}\left({f}\left({x}\right)\right)\:{and}\:{find}\:{its}\:{values}\:. \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{cos}\left(\mathrm{2}{t}\right)}{\left(\mathrm{2}{t}^{\mathrm{2}} \:+{i}\right)^{\mathrm{2}}…
Question Number 186800 by universe last updated on 10/Feb/23 $$\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}} \\ $$$$\int_{\mathrm{0}} ^{\:\infty\:} \frac{\boldsymbol{\mathrm{tan}}^{−\mathrm{1}} \boldsymbol{\alpha{x}}\:\boldsymbol{\mathrm{tan}}^{−\mathrm{1}} \boldsymbol{\beta{x}}}{\boldsymbol{{x}}^{\mathrm{2}} }\boldsymbol{{dx}}\:=\:\frac{\boldsymbol{\pi}}{\mathrm{2}}\mathrm{log}\left\{\frac{\left(\boldsymbol{\alpha}+\boldsymbol{\beta}\right)^{\boldsymbol{\alpha}+\boldsymbol{\beta}} }{\boldsymbol{\alpha}^{\boldsymbol{\alpha}} \boldsymbol{\beta}^{\boldsymbol{\beta}} }\right\}\: \\ $$ Terms of Service…