Question Number 134694 by Engr_Jidda last updated on 06/Mar/21 $$\int{sin}^{\mathrm{4}} {xdx} \\ $$ Answered by john_santu last updated on 06/Mar/21 $$\mathcal{J}\:=\:\int\:\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2}{x}\right)^{\mathrm{2}} {dx} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}}\int\left(\mathrm{1}+\mathrm{cos}\:\mathrm{2}{x}\right)^{\mathrm{2}} \:{dx}…
Question Number 68946 by ramirez105 last updated on 17/Sep/19 Commented by kaivan.ahmadi last updated on 17/Sep/19 $$\frac{\partial{u}}{\partial{x}}=\left(\mathrm{2}{xy}−{tany}\right)\Rightarrow{u}\left({x},{y}\right)={x}^{\mathrm{2}} {y}−{xtany}+{h}\left({y}\right) \\ $$$$\frac{\partial{u}}{\partial{y}}={x}^{\mathrm{2}} −{xsec}^{\mathrm{2}} {y}={x}^{\mathrm{2}} −{xsec}^{\mathrm{2}} {y}+\frac{{dh}}{{dy}}\Rightarrow\frac{{dh}}{{dy}}=\mathrm{0}\Rightarrow{h}={c}' \\…
Question Number 68926 by ramirez105 last updated on 16/Sep/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 68925 by ramirez105 last updated on 16/Sep/19 Commented by kaivan.ahmadi last updated on 16/Sep/19 $$\frac{\partial{u}}{\partial{x}}=\left({x}+\mathrm{3}\right)^{−\mathrm{1}} {cosy}\Rightarrow{u}\left({x},{y}\right)={cosyln}\left({x}+\mathrm{3}\right)+{h}\left({y}\right) \\ $$$$\frac{\partial{u}}{\partial{y}}={y}^{−\mathrm{1}} −{sinyln}\left(\mathrm{5}{x}+\mathrm{15}\right)=−{sinyln}\left({x}+\mathrm{3}\right)+\frac{{dh}}{{dy}}\Rightarrow \\ $$$${y}^{−\mathrm{1}} −{ln}\mathrm{5}=\frac{{dh}}{{dy}}\Rightarrow{h}={lny}−{yln}\mathrm{5} \\…
Question Number 68924 by ramirez105 last updated on 16/Sep/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 68923 by ramirez105 last updated on 16/Sep/19 Commented by kaivan.ahmadi last updated on 16/Sep/19 $$\frac{\partial{u}}{\partial{x}}=\frac{\mathrm{1}}{{y}}\Rightarrow{u}\left({x},{y}\right)=\int\frac{\mathrm{1}}{{y}}{dx}=\frac{{x}}{{y}}+{h}\left({y}\right) \\ $$$$\frac{\partial{u}}{\partial{y}}=−\frac{{x}}{{y}^{\mathrm{2}} }=−\frac{{x}}{{y}^{\mathrm{2}} }+\frac{{dh}}{{dy}}\Rightarrow\frac{{dh}}{{dy}}=\mathrm{0}\Rightarrow{h}={c}' \\ $$$$ \\ $$$$\Rightarrow{u}\left({x},{y}\right)={c}\Rightarrow\frac{{x}}{{y}}={c}\Rightarrow{x}−{cy}=\mathrm{0}…
Question Number 68860 by ramirez105 last updated on 16/Sep/19 Commented by kaivan.ahmadi last updated on 16/Sep/19 $${set}\:{M}={x}+\sqrt{{y}^{\mathrm{2}} +\mathrm{1}}\:\:{and}\:\:\:{N}=−{y}+\frac{{xy}}{\:\sqrt{{y}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$\frac{\partial{M}}{\partial{y}}=\frac{{y}}{\:\sqrt{{y}^{\mathrm{2}} +\mathrm{1}}}=\frac{\partial{N}}{\partial{x}} \\ $$$$\begin{cases}{\frac{\partial{u}}{\partial{x}}={x}+\sqrt{{y}^{\mathrm{2}} +\mathrm{1}}}\\{\frac{\partial{u}}{\partial{y}}=−{y}+\frac{{xy}}{\:\sqrt{{y}^{\mathrm{2}}…
Question Number 68855 by ramirez105 last updated on 16/Sep/19 Commented by kaivan.ahmadi last updated on 16/Sep/19 $${set}\:{M}=\mathrm{6}{x}+{y}^{\mathrm{2}} \:\:{and}\:{N}={y}\left(\mathrm{2}{x}−\mathrm{3}{y}\right) \\ $$$$\frac{\partial{M}}{\partial{y}}=\mathrm{2}{y}\:=\frac{\partial{N}}{\partial{x}}\Rightarrow{the}\:{equation}\:{is}\:{exact} \\ $$$$\frac{\partial{u}}{\partial{x}}=\mathrm{6}{x}+{y}^{\mathrm{2}} \:\:\:{and}\:\:\frac{\partial{u}}{\partial{y}}={y}\left(\mathrm{2}{x}−\mathrm{3}{y}\right)=\mathrm{2}{xy}−\mathrm{3}{y}^{\mathrm{2}} \\ $$$$\Rightarrow{u}\left({x},{y}\right)=\int\left(\mathrm{6}{x}+{y}^{\mathrm{2}}…
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Question Number 68506 by oyemi kemewari last updated on 12/Sep/19 $${y}'=\mathrm{4}{y}^{\mathrm{2}} +{x}^{\mathrm{2}} +\mathrm{1} \\ $$$${what}\:{the}\:{primitive}\:{solution} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com