Question Number 51551 by maxmathsup by imad last updated on 28/Dec/18 $${find}\:{f}\left(\lambda\right)\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \sqrt{\mathrm{1}+\lambda{tant}}{dt}\:\:\:{with}\:\lambda>\mathrm{0} \\ $$$${also}\:{calculate}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{{tant}}{\:\sqrt{\mathrm{1}+\lambda{tant}}}{dt}. \\ $$ Commented by Abdo msup. last…
Question Number 117087 by mnjuly1970 last updated on 09/Oct/20 $$\:\:\:\:\:\:\:\:\:…\:\:{nice}\:\:{calculus}… \\ $$$$\:{please}\:\:{evaluate}\::: \\ $$$$ \\ $$$$\:\:\:\:\:\:\Omega=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{x}^{\mathrm{4}} +\mathrm{1}}{{x}^{\mathrm{6}} +\mathrm{1}}{dx}\:=??? \\ $$$$\:\:\:\:\:\:\:\:\:{m}.{n}.\mathrm{1970} \\ $$ Answered…
Question Number 51550 by maxmathsup by imad last updated on 28/Dec/18 $${find}\:\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}+{x}^{\mathrm{4}} }{dx} \\ $$ Commented by Abdo msup. last updated on 29/Dec/18…
Question Number 117066 by bemath last updated on 09/Oct/20 $$\mathrm{Given}\:\mathrm{f}\left(\mathrm{x}\right)=\:\frac{\mathrm{x}−\mathrm{1}}{\mathrm{x}+\mathrm{1}}\:.\:\mathrm{If}\:\mathrm{f}^{\mathrm{2}} \left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{f}\left(\mathrm{x}\right)\right),\: \\ $$$$\mathrm{f}^{\mathrm{3}} \left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{f}\left(\mathrm{f}\left(\mathrm{x}\right)\right)\right)\:,\:\mathrm{f}^{\mathrm{1998}} \left(\mathrm{x}\right)\:=\:\mathrm{g}\left(\mathrm{x}\right) \\ $$$$\mathrm{then}\:\int_{\frac{\mathrm{1}}{\mathrm{e}}} ^{\mathrm{1}} \mathrm{g}\left(\mathrm{x}\right)\:\mathrm{dx}\:=\:\_? \\ $$ Answered by 1549442205PVT last…
Question Number 51526 by ajfour last updated on 27/Dec/18 Commented by ajfour last updated on 27/Dec/18 $${For}\:{the}\:{area}\:{in}\:{black}\:{and}\:{yellow} \\ $$$${to}\:{be}\:{the}\:{same},\:{find}\:{r}\:{in}\:{terms}\:{of}\:{R}. \\ $$$$ \\ $$ Answered by…
Question Number 117053 by bemath last updated on 09/Oct/20 $$\:\:\:\int\:\frac{\mathrm{x}}{\:\sqrt{\mathrm{2}−\mathrm{x}^{\mathrm{4}} }}\:\mathrm{dx}\:=? \\ $$ Answered by Lordose last updated on 09/Oct/20 $$\mathrm{u}=\mathrm{x}^{\mathrm{2}} \:\Rightarrow\:\mathrm{du}=\mathrm{2xdx} \\ $$$$\mathrm{I}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}−\mathrm{u}^{\mathrm{2}} }}\mathrm{du}…
Question Number 51510 by Tinkutara last updated on 27/Dec/18 Answered by tanmay.chaudhury50@gmail.com last updated on 28/Dec/18 $$\int\frac{−\mathrm{2}{pdp}}{\:\sqrt{{p}^{\mathrm{2}} ×\left(\left\{\mathrm{1}−\mathrm{3}\left({p}^{\mathrm{2}} −\mathrm{1}\right)\right\}\right.}} \\ $$$${x}^{\mathrm{2}} \left\{{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\mathrm{2}\left({x}+\frac{\mathrm{1}}{{x}}\right)−\mathrm{1}\right\} \\…
Question Number 51501 by Tinkutara last updated on 27/Dec/18 Answered by tanmay.chaudhury50@gmail.com last updated on 27/Dec/18 $${x}={tan}\theta\:\:{dx}={sec}^{\mathrm{2}} \theta{d}\theta \\ $$$$\int\frac{\theta}{{tan}^{\mathrm{4}} \theta}{sec}^{\mathrm{2}} \theta{d}\theta \\ $$$$\theta\int\frac{{sec}^{\mathrm{2}} \theta}{{tan}^{\mathrm{4}}…
Question Number 117034 by Lordose last updated on 09/Oct/20 $$\int\frac{\boldsymbol{\mathrm{dx}}}{\:\sqrt[{\mathrm{3}}]{\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{3}} }} \\ $$ Answered by MJS_new last updated on 09/Oct/20 $$\int\frac{{dx}}{\:\sqrt[{\mathrm{3}}]{{x}^{\mathrm{3}} +\mathrm{1}}}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{\sqrt[{\mathrm{3}}]{{x}^{\mathrm{3}} +\mathrm{1}}}{{x}}\:\rightarrow\:{dx}=−{x}^{\mathrm{2}}…
Question Number 117006 by TANMAY PANACEA last updated on 08/Oct/20 $$\int_{\mathrm{0}} ^{\mathrm{100}\pi} \mid{sinx}\mid\:{dx} \\ $$ Commented by TANMAY PANACEA last updated on 08/Oct/20 $${thank}\:{you}\:{sir} \\…