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 $$\Rightarrow\:\frac{\mathrm{1}}{\left({x}−\mathrm{2}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)}\:=\:\frac{{A}}{{x}−\mathrm{2}}\:+\:\frac{{Bx}+{C}}{{x}^{\mathrm{2}} +\mathrm{4}} \\…
Question Number 116813 by bemath last updated on 07/Oct/20 $$\:\:\:\:\:\:\underset{\mathrm{1}} {\overset{\sqrt{\mathrm{3}}} {\int}}\:\frac{\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}{\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx}\:? \\ $$ Answered by bobhans last updated on 07/Oct/20 $$\:\mathrm{letting}\:\mathrm{x}\:=\:\mathrm{tan}\:\theta\:\rightarrow\begin{cases}{\theta=\frac{\pi}{\mathrm{3}}}\\{\theta=\frac{\pi}{\mathrm{4}}}\end{cases} \\…
Question Number 116806 by Backer last updated on 07/Oct/20 $$\mathrm{Hi} \\ $$$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\:\int_{-\infty} ^{+\infty} -\mathrm{e}^{-\mathrm{x}^{\mathrm{2}} } \mathrm{dx}=\sqrt{\pi} \\ $$$$\mathrm{Thanks}\:\mathrm{beforehand} \\ $$$$ \\ $$ Answered…
Question Number 116803 by Ar Brandon last updated on 08/Oct/20 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{X}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right),\:\mathrm{Y}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right),\:\mathrm{Z}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right) \\ $$$$\begin{cases}{\frac{\partial\mathrm{Z}}{\partial\mathrm{y}}−\frac{\partial\mathrm{Y}}{\partial\mathrm{z}}=\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\\{\frac{\partial\mathrm{Z}}{\partial\mathrm{x}}−\frac{\partial\mathrm{X}}{\partial\mathrm{z}}=−\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{2}}}\\{\frac{\partial\mathrm{Y}}{\partial\mathrm{x}}−\frac{\partial\mathrm{X}}{\partial\mathrm{y}}=\mathrm{z}\left(\mathrm{2x}−\mathrm{y}\right)}\end{cases}\:\mathrm{where}\:\begin{cases}{\mathrm{X}\left(\mathrm{x},\mathrm{y},\mathrm{0}\right)=\mathrm{0}}\\{\mathrm{Y}\left(\mathrm{x},\mathrm{y},\mathrm{0}\right)=\mathrm{0}}\\{\mathrm{Z}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)=\mathrm{0}}\end{cases} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com