Question Number 51419 by Tawa1 last updated on 26/Dec/18 $$\int\:\frac{\sqrt{\mathrm{x}}}{\mathrm{1}\:+\:\:\sqrt[{\mathrm{3}}]{\mathrm{x}}}\:\mathrm{dx} \\ $$ Answered by Smail last updated on 27/Dec/18 $${let}\:{u}=\sqrt[{\mathrm{6}}]{{x}}\Rightarrow\mathrm{6}{u}^{\mathrm{5}} {du}={dx} \\ $$$${A}=\int\frac{\sqrt{{x}}}{\mathrm{1}+\sqrt[{\mathrm{3}}]{{x}}}{dx}=\int\frac{{u}^{\mathrm{3}} }{\mathrm{1}+{u}^{\mathrm{2}} }\left(\mathrm{6}{u}^{\mathrm{5}}…
Question Number 51394 by Tawa1 last updated on 26/Dec/18 $$\int\:\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\sqrt{\mathrm{tan}\:\mathrm{x}}}\:\:\mathrm{dx} \\ $$ Answered by peter frank last updated on 26/Dec/18 Commented by Tawa1 last updated…
Question Number 116903 by mnjuly1970 last updated on 07/Oct/20 Answered by mathmax by abdo last updated on 07/Oct/20 $$\mathrm{let}\:\mathrm{f}\left(\mathrm{a}\right)\:=\int_{\mathrm{0}} ^{\pi} \:\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{acosx}\right)}{\mathrm{cosx}}\:\mathrm{dx}\:\:\left(\mathrm{here}\:\mathrm{a}=\mathrm{sin}\alpha\right)\:\Rightarrow \\ $$$$\mathrm{f}^{'} \left(\mathrm{a}\right)\:=\int_{\mathrm{0}} ^{\pi}…
Question Number 51353 by behi83417@gmail.com last updated on 26/Dec/18 Commented by tanmay.chaudhury50@gmail.com last updated on 27/Dec/18 Answered by tanmay.chaudhury50@gmail.com last updated on 26/Dec/18 $$\left.\mathrm{1}\right){I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}}…
Question Number 51324 by Abdo msup. last updated on 25/Dec/18 $${calculate}\:\int_{\mathrm{0}} ^{+\infty} \:\:\:\:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}\:} \:+{x}^{\mathrm{4}} } \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 26/Dec/18 $$\int\frac{{dx}}{{x}^{\mathrm{4}}…
Question Number 116846 by mnjuly1970 last updated on 07/Oct/20 $$\:\:\:\:\:\:\:\:\:\:\:\:\:…{nice}\:\:{calculus}… \\ $$$$\:\:\:{prove}\:\:{that}\::: \\ $$$$ \\ $$$$\:\: \\ $$$$\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \sqrt{\frac{\left(\mathrm{2}^{{x}} −\mathrm{1}\right){sin}^{\mathrm{3}} \left({x}\right)}{\left(\mathrm{2}^{{x}} +\mathrm{1}\right)\left({sin}^{\mathrm{3}} \left({x}\right)+{cos}^{\mathrm{3}} \left({x}\right)\right)}}\:\:{dx}<\frac{\pi}{\mathrm{8}}…
Question Number 116844 by bemath last updated on 07/Oct/20 $$\int\:\frac{\mathrm{8x}+\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{2x}\right)}{\:\sqrt{\mathrm{1}−\mathrm{4x}^{\mathrm{2}} }}\:\mathrm{dx}\: \\ $$ Answered by Dwaipayan Shikari last updated on 07/Oct/20 $$\int\frac{\mathrm{8}{x}}{\:\sqrt{\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} }}+\int\frac{{sin}^{−\mathrm{1}} \left(\mathrm{2}{x}\right)}{\:\sqrt{\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}}…
Question Number 182368 by universe last updated on 08/Dec/22 $$\:\:\:\:\mathrm{find}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{region}\:\mathrm{in}\:\:\mathbb{R}^{\mathrm{3}} \:\:\mathrm{given}\:\mathrm{by}\: \\ $$$$\:\:\:\:\mathrm{3}\mid\mathrm{x}\mid\:+\:\mathrm{4}\mid\mathrm{y}\mid\:+\mathrm{3}\mid\mathrm{z}\mid\:\leqslant\mathrm{12}\:\:\mathrm{is} \\ $$ Answered by mr W last updated on 08/Dec/22 $$\frac{\mid{x}\mid}{\mathrm{4}}+\frac{\mid{y}\mid}{\mathrm{3}}+\frac{\mid{z}\mid}{\mathrm{4}}\leqslant\mathrm{1} \\…
Question Number 182367 by universe last updated on 08/Dec/22 $$ \\ $$$$\:\:\:\:\mathrm{find}\:\mathrm{volume}\:\mathrm{of}\:\:\mathrm{region}\:\mathrm{bounded}\:\mathrm{above}\: \\ $$$$\:\:\:\mathrm{by}\:\mathrm{z}\:=\:\mathrm{1}+\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} \:}\:\:\mathrm{and}\:\mathrm{below} \\ $$$$\:\:\:\:\mathrm{by}\:\:\:\mathrm{z}\:=\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \:}\: \\ $$ Answered by Tokugami…
Question Number 116815 by bemath last updated on 07/Oct/20 $$\int\:\frac{\mathrm{dx}}{\left(\mathrm{x}−\mathrm{2}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{4}\right)}\:=? \\ $$ Answered by john santu last updated on 07/Oct/20 $$\Rightarrow\:\frac{\mathrm{1}}{\left({x}−\mathrm{2}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)}\:=\:\frac{{A}}{{x}−\mathrm{2}}\:+\:\frac{{Bx}+{C}}{{x}^{\mathrm{2}} +\mathrm{4}} \\…