Question Number 131068 by EDWIN88 last updated on 01/Feb/21 $$\:\int\:\mathrm{sin}\:^{\mathrm{2}} {x}\:\mathrm{cos}\:\mathrm{4}{x}\:{dx}\:=?\: \\ $$ Answered by Ar Brandon last updated on 01/Feb/21 $$\mathcal{I}=\int\mathrm{sin}^{\mathrm{2}} \mathrm{xcos4xdx}=\frac{\mathrm{1}}{\mathrm{2}}\int\left(\mathrm{1}+\mathrm{cos2x}\right)\mathrm{cos4xdx} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int\left[\mathrm{cos4x}+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{cos6x}+\mathrm{cos2x}\right)\right]\mathrm{dx}…
Question Number 131058 by pipin last updated on 01/Feb/21 $$\int\frac{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{2}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{3}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{4}\right)}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{5}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{6}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{7}\right)}\:\mathrm{dx}\: \\ $$ Answered by MJS_new last updated on 01/Feb/21…
Question Number 131053 by pipin last updated on 01/Feb/21 $$\int\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{2}}{\mathrm{x}^{\mathrm{4}} +\mathrm{4}}\mathrm{dx} \\ $$ Answered by Ar Brandon last updated on 01/Feb/21 $$\mathcal{I}=\int\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{2}}{\mathrm{x}^{\mathrm{4}} +\mathrm{4}}\mathrm{dx}=\int\frac{\mathrm{1}+\frac{\mathrm{2}}{\mathrm{x}^{\mathrm{2}}…
Question Number 131050 by mathmax by abdo last updated on 31/Jan/21 $$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{tsin}\left(\mathrm{xt}\right)}{\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\mathrm{dt}\:\:\:\left(\mathrm{x}>\mathrm{0}\right) \\ $$$$\mathrm{calculate}\:\mathrm{f}^{'} \left(\mathrm{x}\right) \\ $$ Answered by mathmax…
Question Number 131049 by mathmax by abdo last updated on 31/Jan/21 $$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{cos}\left(\mathrm{xt}\right)}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{t}^{\mathrm{2}} }\mathrm{dt}\:\:\mathrm{calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx} \\ $$ Answered by mindispower last updated…
Question Number 65486 by aliesam last updated on 30/Jul/19 Answered by MJS last updated on 31/Jul/19 $$\int\mathrm{arctan}\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}\:{dx}= \\ $$$$\:\:\:\:\:{u}'=\mathrm{1}\:\rightarrow\:{u}={x} \\ $$$$\:\:\:\:\:{v}=\mathrm{arctan}\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}\:\rightarrow\:{v}'=−\frac{\mathrm{2}{x}−\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)}…
Question Number 65485 by aliesam last updated on 30/Jul/19 $$\int{e}^{{cos}^{−\mathrm{1}} \left({x}\right)} \:{dx} \\ $$ Answered by MJS last updated on 31/Jul/19 $$\int\mathrm{e}^{\mathrm{arccos}\:{x}} {dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{arccos}\:{x}\:\rightarrow\:{dx}=−\mathrm{sin}\:{t}\:{dt}\right]…
Question Number 65478 by smz last updated on 30/Jul/19 Commented by MJS last updated on 31/Jul/19 $$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{it}'\mathrm{s}\:\mathrm{possible} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 65455 by mathmax by abdo last updated on 30/Jul/19 $${find}\:{U}_{{n}} =\:\int_{\mathrm{0}} ^{+\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{x}} }{\mathrm{2}^{{x}^{\mathrm{2}} } \left({x}^{\mathrm{2}} \:+\mathrm{4}{n}^{\mathrm{2}} \right)}{dx}\:\:\:\:\:\left({n}\:{from}\:{N}\:{and}\:{n}\geqslant\mathrm{1}\right) \\ $$$${study}\:{nature}\:{of}\:{the}\:{serie}\:\:\Sigma\:\mathrm{2}^{{n}^{\mathrm{2}} } {U}_{{n}} \\…
Question Number 65450 by imron876 last updated on 30/Jul/19 Commented by Prithwish sen last updated on 30/Jul/19 $$\left.\mathrm{1}\right)\:\mathrm{a}=\mathrm{b}=\mathrm{c}=\mathrm{d}=\mathrm{1} \\ $$$$\left.\mathrm{2}\right)\mathrm{a}=\mathrm{c}=\mathrm{1},\mathrm{b}=\mathrm{d}=\mathrm{0} \\ $$$$\left.\mathrm{3}\right)\mathrm{b}=\mathrm{d}=\mathrm{1},\mathrm{a}=\mathrm{c}=\mathrm{0} \\ $$$$\left.\mathrm{4}\right)\mathrm{a}=\mathrm{1}\:\mathrm{c}=−\mathrm{1}\:\mathrm{b}=\mathrm{d}=\mathrm{0} \\…