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…
Question Number 54919 by maxmathsup by imad last updated on 14/Feb/19 $${calculate}\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \left({x}^{\mathrm{3}} +\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}{dx} \\ $$ Commented by Abdo msup. last updated on…
Question Number 185985 by cortano1 last updated on 30/Jan/23 $$\:\:\:\int\:\frac{{dx}}{\:\sqrt{\mathrm{sin}\:{x}\left(\mathrm{1}+\mathrm{cos}\:{x}\right)}}\:=? \\ $$ Answered by ARUNG_Brandon_MBU last updated on 30/Jan/23 $${I}=\int\frac{{dx}}{\:\sqrt{\mathrm{sin}{x}\left(\mathrm{1}+\mathrm{cos}{x}\right)}}=\int\frac{{dx}}{\mathrm{2}\sqrt{\mathrm{sin}\frac{{x}}{\mathrm{2}}\mathrm{cos}^{\mathrm{3}} \frac{{x}}{\mathrm{2}}}} \\ $$$$\:\:=\int\frac{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{\:\sqrt{\mathrm{tan}\frac{{x}}{\mathrm{2}}}}{dx}=\int\frac{{d}\left(\mathrm{tan}\frac{{x}}{\mathrm{2}}\right)}{\:\sqrt{\mathrm{tan}\frac{{x}}{\mathrm{2}}}}=\mathrm{2}\sqrt{\mathrm{tan}\frac{{x}}{\mathrm{2}}}+{C} \\…
Question Number 120438 by bramlexs22 last updated on 31/Oct/20 Answered by TANMAY PANACEA last updated on 31/Oct/20 $${formula} \\ $$$$\int_{{a}} ^{{b}} {f}\left({x}\right){dx}=\int_{\mathrm{0}} ^{{b}} {f}\left({a}+{b}−{x}\right){dx} \\…