Question Number 120554 by snipers237 last updated on 01/Nov/20 $${Prove}\:{that}\:{for}\:{all}\:\:{a}>\mathrm{0} \\ $$$$\int_{\left[−{a};{a}\right]} {arg}\left(\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\:−{ix}\right)\right){dx}\:=\mathrm{0} \\ $$$${Deduce}\:{that}\: \\ $$$${f}:\:{x}\rightarrow{arg}\left(\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\:−{ix}\right)\right)\:\:{is}\:{an}\:{old}\:{function}\:{on}\:\mathbb{R} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 120552 by talminator2856791 last updated on 01/Nov/20 $$\: \\ $$$$\: \\ $$$$\:\:\:\:\:\:\:\mathrm{evaluate}:\:\int_{\mathrm{0}} ^{\:\infty} \left(\frac{{x}}{{x}+\mathrm{1}}\right)^{{x}!} {dx} \\ $$$$ \\ $$$$ \\ $$ Terms of…
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Question Number 120549 by bobhans last updated on 01/Nov/20 Answered by TANMAY PANACEA last updated on 01/Nov/20 $$\frac{−\mathrm{1}}{\mathrm{4}}\int\frac{−\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{3}−\mathrm{4}{x}−\mathrm{3}}{\left(\mathrm{3}+\mathrm{4}{x}−\mathrm{4}{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{dx} \\ $$$$=\frac{−\mathrm{1}}{\mathrm{4}}\int\frac{{dx}}{\:\sqrt{\mathrm{3}+\mathrm{4}{x}−\mathrm{4}{x}^{\mathrm{2}} }}+\frac{\mathrm{1}}{\mathrm{4}}\int\frac{\mathrm{4}{x}+\mathrm{3}}{\left(\mathrm{3}+\mathrm{4}{x}−\mathrm{4}{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}}…
Question Number 120544 by rubygarfield last updated on 01/Nov/20 $$\int\frac{\mathrm{1}}{\left(\mathrm{cos}\:{x}\right)^{\mathrm{6}} }=? \\ $$ Answered by Dwaipayan Shikari last updated on 01/Nov/20 $$\int{sec}^{\mathrm{6}} {xdx} \\ $$$$=\int{sec}^{\mathrm{2}}…
Question Number 54995 by peter frank last updated on 15/Feb/19 $$\int\:\sqrt[{\mathrm{3}}]{{x}^{\mathrm{3}} +{x}^{\mathrm{2}} }\:{dx} \\ $$ Commented by MJS last updated on 16/Feb/19 $$\mathrm{I}\:\mathrm{can}\:\mathrm{solve}\:\mathrm{this}\:\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\mathrm{a}\:\mathrm{lot}\:\mathrm{of}\:\mathrm{typing}\:\mathrm{work} \\ $$$$\mathrm{if}\:\mathrm{you}\:\mathrm{need}\:\mathrm{it}\:\mathrm{urgently}\:\mathrm{I}\:\mathrm{will}\:\mathrm{post}\:\mathrm{it}…
Question Number 186048 by ARUNG_Brandon_MBU last updated on 31/Jan/23 $$\int_{{a}} ^{{b}} \frac{{dx}}{\:\sqrt{{b}−{x}}+\sqrt{{x}−{a}}} \\ $$ Commented by ARUNG_Brandon_MBU last updated on 31/Jan/23 $$\neq\mathrm{0} \\ $$ Commented…
Question Number 54972 by peter frank last updated on 15/Feb/19 $$\int_{\mathrm{0}} ^{\pi} \int_{\mathrm{0}} ^{\mathrm{4cos}\:{z}} \int_{\mathrm{0}} ^{\sqrt{\mathrm{16}−{y}^{\mathrm{2}} }} {ydxdydz} \\ $$ Answered by kaivan.ahmadi last updated…
Question Number 54971 by peter frank last updated on 15/Feb/19 $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\sqrt{{tanx}}\:\mathrm{sin}\:^{\mathrm{2}} {x}}{\:\sqrt{\mathrm{sin}\:{x}\mathrm{cos}\:{x}}\:\mathrm{tan}\:{x}}{dx} \\ $$ Answered by kaivan.ahmadi last updated on 15/Feb/19 $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}}…
Question Number 54936 by maxmathsup by imad last updated on 14/Feb/19 $${let}\:{f}\left(\theta\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\sqrt{{x}^{\mathrm{2}} \:+\mathrm{2}\left({cos}\theta\right){x}\:+\mathrm{1}}{dx}\:\:\:{with}\:\theta\:\in\:{R}\:. \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}\left(\theta\right) \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:{g}\left(\theta\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{xsin}\theta}{\:\sqrt{{x}^{\mathrm{2}} \:+\mathrm{2}{cos}\theta\:{x}\:+\mathrm{1}}}{dx} \\ $$ Commented…