Question Number 129859 by EDWIN88 last updated on 20/Jan/21 $$\:\mathrm{L}\:=\:\int_{−\mathrm{1}} ^{\:\mathrm{0}} \sqrt{\frac{\mathrm{1}+\mathrm{y}}{\mathrm{1}−\mathrm{y}}}\:\mathrm{dy}\: \\ $$ Answered by liberty last updated on 20/Jan/21 $$\:\mathrm{let}\:\mathrm{y}=\mathrm{cos}\:\mathrm{2t}\:\rightarrow\begin{cases}{\mathrm{y}=\mathrm{0}\rightarrow\mathrm{t}=\frac{\pi}{\mathrm{4}}}\\{\mathrm{y}=−\mathrm{1}\rightarrow\mathrm{t}=\frac{\pi}{\mathrm{2}}}\end{cases} \\ $$$$\mathrm{L}\:=\int_{\pi/\mathrm{2}} ^{\:\pi/\mathrm{4}}…
Question Number 129856 by mohammad17 last updated on 20/Jan/21 Commented by mohammad17 last updated on 20/Jan/21 $${how}\:{can}\:{solve}\:{this} \\ $$ Answered by MJS_new last updated on…
Question Number 64320 by aliesam last updated on 16/Jul/19 $${without}\:{beta}\:{function} \\ $$$$\int{cos}^{\mathrm{3}} {t}\:{sin}^{\mathrm{2}} {t}\:{dt}\: \\ $$ Commented by mathmax by abdo last updated on 16/Jul/19…
Question Number 129855 by liberty last updated on 20/Jan/21 $$\:\vartheta\:=\:\int\:\frac{{dx}}{\left(\mathrm{1}+\sqrt{{x}}\:\right)^{\mathrm{3}} } \\ $$ Answered by EDWIN88 last updated on 20/Jan/21 $$\vartheta\:=\:\int\:\frac{\mathrm{dx}}{\left(\sqrt{\mathrm{x}}\right)^{\mathrm{3}} \left(\mathrm{1}+\mathrm{x}^{−\mathrm{1}/\mathrm{2}} \right)^{\mathrm{3}} }=\:\int\:\frac{\mathrm{x}^{−\mathrm{3}/\mathrm{2}} }{\left(\mathrm{1}+\mathrm{x}^{−\mathrm{1}/\mathrm{2}}…
Question Number 64311 by aseer imad last updated on 16/Jul/19 Commented by Tony Lin last updated on 17/Jul/19 $${letA}'={A}−\left(\mathrm{1},\mathrm{3}\right)=\left(\mathrm{0},\mathrm{0}\right),{C}'={C}−\left(\mathrm{1},\mathrm{3}\right)=\left(\mathrm{4},\mathrm{5}\right) \\ $$$${D}'=\begin{bmatrix}{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\end{bmatrix}\begin{bmatrix}{{cos}\frac{\pi}{\mathrm{4}}\:\:−{sin}\frac{\pi}{\mathrm{4}}}\\{{sin}\frac{\pi}{\mathrm{4}}\:\:\:\:\:\:{cos}\frac{\pi}{\mathrm{4}}}\end{bmatrix}\begin{bmatrix}{\mathrm{4}}\\{\mathrm{5}}\end{bmatrix} \\ $$$$\:\:\:\:\:\:\:=\begin{bmatrix}{\frac{\mathrm{1}}{\mathrm{2}}\:\:−\frac{\mathrm{1}}{\mathrm{2}}}\\{\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}}\end{bmatrix}\begin{bmatrix}{\mathrm{4}}\\{\mathrm{5}}\end{bmatrix} \\ $$$$\:\:\:\:\:\:\:\:=\begin{bmatrix}{−\frac{\mathrm{1}}{\mathrm{2}}}\\{\:\:\:\:\frac{\mathrm{9}}{\mathrm{2}}}\end{bmatrix}…
Question Number 64309 by aseer imad last updated on 16/Jul/19 Commented by Tony Lin last updated on 17/Jul/19 $${let}\:{A}'={A}−\left(\mathrm{1},\mathrm{3}\right)=\left(\mathrm{0},\mathrm{0}\right),{C}'={C}−\left(\mathrm{1},\mathrm{3}\right)=\left(\mathrm{4},\mathrm{5}\right) \\ $$$${D}'\Rightarrow\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{4}+\mathrm{5}{i}\right)\left({cos}\frac{\pi}{\mathrm{4}}+{isin}\frac{\pi}{\mathrm{4}}\right)=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{9}}{\mathrm{2}}{i} \\ $$$$\Rightarrow{D}=\left(\mathrm{1},\mathrm{3}\right)+\left(−\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{9}}{\mathrm{2}}\right)=\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{15}}{\mathrm{2}}\right) \\ $$$${B}'\Rightarrow\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{4}+\mathrm{5}{i}\right)\left({cos}\frac{\pi}{\mathrm{4}}−{isin}\frac{\pi}{\mathrm{4}}\right)=\frac{\mathrm{9}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}{i}…
Question Number 64305 by aliesam last updated on 16/Jul/19 Commented by aliesam last updated on 16/Jul/19 $${prove}\:{that} \\ $$ Commented by mathmax by abdo last…
Question Number 129841 by Adel last updated on 20/Jan/21 $$\left\{_{\sqrt[{\mathrm{3}}]{\mathrm{2}^{\mathrm{2x}} }+\sqrt[{\mathrm{3}}]{\mathrm{2}^{\mathrm{4x}} }+\mathrm{4}^{\mathrm{x}} =\mathrm{5}} ^{\sqrt[{\mathrm{3}}]{\mathrm{2}^{\mathrm{x}} \:}−\mathrm{2}^{\mathrm{x}} =\mathrm{2}} \right\}\Rightarrow\mathrm{2}^{\mathrm{x}} −\mathrm{8}^{\mathrm{x}} =? \\ $$$$\mathrm{pleas}\:\mathrm{solve}\:\mathrm{thes} \\ $$ Commented by…
Question Number 129839 by liberty last updated on 20/Jan/21 Answered by EDWIN88 last updated on 20/Jan/21 $$\mathrm{J}=\int_{\:\mathrm{0}} ^{\:\pi/\mathrm{2}} \:\frac{\mathrm{3}\sqrt{\mathrm{cos}\:\mathrm{x}}}{\left(\sqrt{\mathrm{cos}\:\mathrm{x}}\:+\sqrt{\mathrm{sin}\:\mathrm{x}}\:\right)^{\mathrm{5}} }\:\mathrm{dx}\: \\ $$$$\:\mathrm{let}\:\mathrm{x}=\frac{\pi}{\mathrm{2}}−\mathrm{t}\:\Rightarrow\mathrm{J}=\int_{\frac{\pi}{\mathrm{2}}} ^{\:\mathrm{0}} \:\frac{\mathrm{3}\sqrt{\mathrm{sin}\:\mathrm{t}}}{\left(\sqrt{\mathrm{sin}\:\mathrm{t}}\:+\sqrt{\mathrm{cos}\:\mathrm{t}}\right)^{\mathrm{5}} }\left(−\mathrm{dt}\right)…
Question Number 64302 by mathmax by abdo last updated on 16/Jul/19 Terms of Service Privacy Policy Contact: info@tinkutara.com