Question Number 7263 by Tawakalitu. last updated on 19/Aug/16 Commented by Rasheed Soomro last updated on 20/Aug/16 $${I}\:{have}\:{answered}\:{your}\:{question}#\mathrm{6852} \\ $$$${which}\:{is}\:{very}\:{resembling}\:{to}\:{this}\:{question}. \\ $$$${Please}\:{see}\:{if}\:{you}\:{have}\:{not}\:{seen}\:{yet}. \\ $$ Commented…
Question Number 138335 by Ñï= last updated on 12/Apr/21 $$\int\frac{{dx}}{{x}^{\mathrm{8}} +{x}^{\mathrm{4}} +\mathrm{1}}=? \\ $$ Answered by MJS_new last updated on 12/Apr/21 $${x}^{\mathrm{8}} +{x}^{\mathrm{4}} +\mathrm{1}= \\…
Question Number 72796 by Learner-123 last updated on 03/Nov/19 $${Integrate}\:{f}\left({x},{y}\right)=\frac{\mathrm{1}}{\left(\mathrm{1}+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{over} \\ $$$${the}\:{triangle}\:{with}\:{vertices}\:\left(\mathrm{0},\mathrm{0}\right)\:,\left(\mathrm{1},\mathrm{0}\right), \\ $$$$\left(\mathrm{1},\sqrt{\mathrm{3}}\right)\:{after}\:{changing}\:{it}\:{to}\:{polar}\:{form}. \\ $$ Answered by mind is power last…
Question Number 138329 by JulioCesar last updated on 12/Apr/21 Answered by Ñï= last updated on 12/Apr/21 $$\int\frac{{dx}}{\mathrm{sin}\:^{\mathrm{4}} {x}+\mathrm{cos}\:^{\mathrm{4}} {x}}=\int\frac{{dx}}{\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} {x}\mathrm{cos}\:^{\mathrm{2}} {x}}=\int\frac{{dx}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}{x}} \\ $$$$=\int\frac{{csc}^{\mathrm{2}} \mathrm{2}{xd}}{{csc}^{\mathrm{2}}…
Question Number 7258 by Tawakalitu. last updated on 19/Aug/16 Commented by Yozzia last updated on 19/Aug/16 $${y}'=\frac{{x}}{\:\sqrt{\mathrm{5}−{x}^{\mathrm{2}} }}\Rightarrow{y}=\int\frac{{x}}{\:\sqrt{\mathrm{5}−{x}^{\mathrm{2}} }}{dx}. \\ $$$${Let}\:{u}=\mathrm{5}−{x}^{\mathrm{2}} \Rightarrow{du}=−\mathrm{2}{xdx}\Rightarrow{xdx}=\frac{−\mathrm{1}}{\mathrm{2}}{du}. \\ $$$${y}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{−{du}}{\:\sqrt{{u}}}=\frac{−\mathrm{1}}{\mathrm{2}}\int{u}^{−\mathrm{1}/\mathrm{2}} {du}=\frac{−\mathrm{1}}{\mathrm{2}}×\mathrm{2}{u}^{\mathrm{1}/\mathrm{2}}…
Question Number 138330 by JulioCesar last updated on 12/Apr/21 Answered by Dwaipayan Shikari last updated on 12/Apr/21 $${tanh}^{−\mathrm{1}} {a}=\frac{\mathrm{1}}{\mathrm{2}}{log}\left(\frac{\mathrm{1}+{a}}{\mathrm{1}−{a}}\right) \\ $$$${tanh}^{−\mathrm{1}} {a}+{tanh}^{−\mathrm{1}} {b}=\frac{\mathrm{1}}{\mathrm{2}}{log}\left(\frac{\left(\mathrm{1}+{a}\right)\left(\mathrm{1}+{b}\right)}{\left(\mathrm{1}−{a}\right)\left(\mathrm{1}−{b}\right)}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{log}\left(\frac{\mathrm{1}+{ab}+{a}+{b}}{\mathrm{1}+{ab}−{a}−{b}}\right)…
Question Number 72789 by Learner-123 last updated on 02/Nov/19 $${Find}\:{the}\:{area}\:{of}\:{the}\:{surface}\:{generated} \\ $$$${by}\:{revolving}\:{the}\:{curve}\:{x}=\frac{{y}^{\mathrm{4}} }{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{8}{y}^{\mathrm{2}} }\: \\ $$$${about}\:{the}\:{x}−{axis}\:.\:\left({given}:\mathrm{1}\leqslant{y}\leqslant\mathrm{2}\right) \\ $$ Commented by MJS last updated on 02/Nov/19…
Question Number 7251 by Tawakalitu. last updated on 19/Aug/16 $${Find}\:{the}\:{value}\:{of}\:{x}\: \\ $$$$ \\ $$$$\left(\sqrt{\mathrm{2}\:+\:\sqrt{\mathrm{3}}}\right)^{{x}} \:\:+\:\:\left(\sqrt{\mathrm{2}\:−\:\sqrt{\mathrm{3}}}\right)^{{x}} \:\:=\:\:\mathrm{4} \\ $$ Commented by sou1618 last updated on 19/Aug/16…
Question Number 138320 by aliibrahim1 last updated on 12/Apr/21 Answered by Rasheed.Sindhi last updated on 17/Apr/21 $$\mathrm{11}^{{n}} =\left(\mathrm{1}+\mathrm{10}\right)^{{n}} =\begin{pmatrix}{{n}}\\{\mathrm{0}}\end{pmatrix}\left(\mathrm{10}\right)^{\mathrm{0}} +\begin{pmatrix}{{n}}\\{\mathrm{1}}\end{pmatrix}\left(\mathrm{10}\right)^{\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:+\begin{pmatrix}{{n}}\\{\mathrm{2}}\end{pmatrix}\left(\mathrm{10}\right)^{\mathrm{2}} +\underset{{divisible}\:{by}\:\mathrm{1000}} {\underbrace{\begin{pmatrix}{{n}}\\{\mathrm{3}}\end{pmatrix}\left(\mathrm{10}\right)^{\mathrm{3}} ….+\begin{pmatrix}{{n}}\\{{n}}\end{pmatrix}\left(\mathrm{10}\right)^{{n}}…
Question Number 72787 by Maclaurin Stickker last updated on 02/Nov/19 $${If}\:{z}_{\mathrm{1}} =\mathrm{6}\left(\mathrm{cos}\:\frac{\pi}{\mathrm{4}}+\mathrm{sin}\:\frac{\pi}{\mathrm{4}}\right)\:{and} \\ $$$${z}_{\mathrm{2}} =\mathrm{2}\left(\mathrm{cos}\:\frac{\pi}{\mathrm{5}}+\mathrm{i}×\mathrm{sin}\:\frac{\pi}{\mathrm{5}}\right)\:{calculate}\:\frac{{z}_{\mathrm{1}} }{{z}_{\mathrm{2}} }. \\ $$ Commented by MJS last updated on…