# cos-2-ln-cos-sin-cos-sin-

Question Number 76963 by peter frank last updated on 01/Jan/20
$$\int\mathrm{cos}\:\mathrm{2}\theta\:\mathrm{ln}\:\left(\frac{\mathrm{cos}\:\theta+\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta−\mathrm{sin}\:\theta}\right) \\$$
Answered by MJS last updated on 02/Jan/20
$$\int\mathrm{cos}\:\mathrm{2}\theta\:\mathrm{ln}\:\frac{\mathrm{cos}\:\theta\:+\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta\:−\mathrm{sin}\:\theta}\:{d}\theta= \\$$$$=\int\mathrm{cos}\:\mathrm{2}\theta\:\mathrm{ln}\:\frac{\mathrm{cos}\:\mathrm{2}\theta}{\mathrm{1}−\mathrm{sin}\:\mathrm{2}\theta}\:{d}\theta= \\$$$$\:\:\:\:\:\mathrm{by}\:\mathrm{parts} \\$$$$\:\:\:\:\:{u}=\mathrm{ln}\:\frac{\mathrm{cos}\:\mathrm{2}\theta}{\mathrm{1}−\mathrm{sin}\:\mathrm{2}\theta}\:\rightarrow\:{u}'=\frac{\mathrm{2}}{\mathrm{cos}\:\mathrm{2}\theta} \\$$$$\:\:\:\:\:{v}'=\mathrm{cos}\:\mathrm{2}\theta\:\rightarrow\:{v}=\frac{\mathrm{sin}\:\mathrm{2}\theta}{\mathrm{2}} \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{2}\theta\:\mathrm{ln}\:\frac{\mathrm{cos}\:\mathrm{2}\theta}{\mathrm{1}−\mathrm{sin}\:\mathrm{2}\theta}\:−\int\mathrm{tan}\:\mathrm{2}\theta\:{d}\theta= \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{2}\theta\:\mathrm{ln}\:\frac{\mathrm{cos}\:\mathrm{2}\theta}{\mathrm{1}−\mathrm{sin}\:\mathrm{2}\theta}\:−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mathrm{cos}\:\mathrm{2}\theta\:= \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{sin}\:\mathrm{2}\theta\:\mathrm{ln}\:\frac{\mathrm{cos}\:\mathrm{2}\theta}{\mathrm{1}−\mathrm{sin}\:\mathrm{2}\theta}\:−\mathrm{ln}\:\mathrm{cos}\:\mathrm{2}\theta\right)\:+{C} \\$$
Commented by john santu last updated on 02/Jan/20
Commented by john santu last updated on 02/Jan/20
$${to}\:{Mr}\:{MJS}\: \\$$
Commented by peter frank last updated on 02/Jan/20
$${thank}\:{you} \\$$
Commented by MJS last updated on 02/Jan/20
$$\mathrm{I}\:\mathrm{get}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}=\frac{\mathrm{1}}{\mathrm{2}} \\$$$$\mathrm{by}\:\mathrm{calculating}\:{f}^{−\mathrm{1}} \left({x}\right)\:\mathrm{and}\:\mathrm{then}\:\mathrm{approximating} \\$$
Commented by jagoll last updated on 02/Jan/20
$$\mathrm{how}\:\mathrm{sir}?\:\mathrm{please}\:\mathrm{your}\:\mathrm{write} \\$$$$\\$$
Commented by MJS last updated on 02/Jan/20
$${y}=\mathrm{8}{x}^{\mathrm{3}} +\mathrm{3}{x} \\$$$$\mathrm{exchanging}\:{x}\rightleftharpoons{y} \\$$$${y}^{\mathrm{3}} +\frac{\mathrm{3}}{\mathrm{8}}{y}−\frac{{x}}{\mathrm{8}}=\mathrm{0} \\$$$${D}=\frac{{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{{q}^{\mathrm{2}} }{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{512}}+\frac{{x}^{\mathrm{2}} }{\mathrm{256}}\geqslant\mathrm{0}\forall{y}\in\mathbb{R}\:\Rightarrow\:\mathrm{Cardano}'\mathrm{s}\:\mathrm{firmula} \\$$$$\Rightarrow \\$$$${f}^{−\mathrm{1}} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{4}}}\left(\sqrt[{\mathrm{3}}]{\mathrm{2}{x}+\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{2}}}−\sqrt[{\mathrm{3}}]{−\mathrm{2}{x}+\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{2}}}\right) \\$$$${g}\left({x}\right)=\frac{{f}^{−\mathrm{1}} \left(\mathrm{8}{x}\right)−{f}^{−\mathrm{1}} \left({x}\right)}{{x}^{\mathrm{1}/\mathrm{3}} } \\$$$${g}\left(\mathrm{10}^{\mathrm{1}} \right)=.\mathrm{526706}… \\$$$${g}\left(\mathrm{10}^{\mathrm{2}} \right)=.\mathrm{505799}… \\$$$${g}\left(\mathrm{10}^{\mathrm{3}} \right)=.\mathrm{501249}… \\$$$${g}\left(\mathrm{10}^{\mathrm{4}} \right)=.\mathrm{500269}… \\$$$${g}\left(\mathrm{10}^{\mathrm{5}} \right)=.\mathrm{500058}… \\$$$$… \\$$
Commented by MJS last updated on 02/Jan/20
$$…\mathrm{I}\:\mathrm{do}\:\mathrm{not}\:\mathrm{understand}\:\mathrm{your}\:\mathrm{method}… \\$$
Commented by jagoll last updated on 02/Jan/20
$$\mathrm{sorry}\:\mathrm{sir}.\:\mathrm{by}\:\mathrm{calculate}\:\mathrm{i}\:\mathrm{got}\:\mathrm{result} \\$$$$\frac{\sqrt[{\mathrm{3}}]{\mathrm{4}}}{\mathrm{2}}\:\mathrm{sir}=\:\mathrm{not}\:\mathrm{same}\:\frac{\mathrm{1}}{\mathrm{2}} \\$$
Commented by mr W last updated on 02/Jan/20
$${to}\:{john}\:{santu}\:{sir}: \\$$$${your}\:{step}\:{to} \\$$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left[\frac{\mathrm{8}}{\mathrm{24}\left({g}\left({x}\right)\right)^{\mathrm{2}} +\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{24}\left({h}\left({x}\right)\right)^{\mathrm{2}} ×\mathrm{3}}\right]×\mathrm{3}{x}^{\frac{\mathrm{2}}{\mathrm{3}}} \\$$$${is}\:{correct}.\:{but}\:{this}\:{doesn}'{t}\:{help}\:{you} \\$$$${further},\:{since}\:{g}\left({x}\right)={f}^{\:−\mathrm{1}} \left(\mathrm{8}{x}\right)\:{and} \\$$$${h}\left({x}\right)={f}^{\:−\mathrm{1}} \left({x}\right)\:{are}\:{still}\:{unknown}. \\$$$${your}\:{last}\:{step}\:{to} \\$$$$=\mathrm{3}×\left[\frac{\mathrm{8}}{\mathrm{24}}−\frac{\mathrm{1}}{\mathrm{24}}\right] \\$$$${is}\:{wrong}.\:{how}\:{can}\:{you}\:{get}\:{this}\:{if} \\$$$${you}\:{don}'{t}\:{explictly}\:{know}\:{g}\left({x}\right)\:{and} \\$$$${h}\left({x}\right)? \\$$
Commented by john santu last updated on 02/Jan/20
$${ok}\:{sir},\:{i}\:{agree}.\:{i}\:{thought}\:{the}\:{degrees}\: \\$$$${g}\left({x}\right)\:{and}\:{h}\left({x}\right)\:{were}\:{the}\:{same},\:{but}\: \\$$$${forgot}\:{the}\:{coefficients}\:{that}\:{might}\: \\$$$${not}\:{be}\:{equal}\:{to}\:\mathrm{1}.\: \\$$
Commented by mr W last updated on 02/Jan/20
$${the}\:{degrees}\:{of}\:{g}\left({x}\right)\:{and}\:{h}\left({x}\right)\:{could} \\$$$${be}\:{the}\:{same},\:{but}\:{you}\:{have}\:{here} \\$$$$\left({g}\left({x}\right)\right)^{\mathrm{2}} \:{and}\:\left({h}\left({x}\right)\right)^{\mathrm{2}} ,\:{and}\:{we}\:{don}'{t} \\$$$${even}\:{know}\:{the}\:{degrees}\:{of}\:{them}. \\$$$${therefore}\:{we}\:{don}'{t}\:{know}\:{the}\:{values}\:{of} \\$$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\left({g}\left({x}\right)\right)^{\mathrm{2}} }{{x}^{\frac{\mathrm{2}}{\mathrm{3}}} }\:{and}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\left({h}\left({x}\right)\right)^{\mathrm{2}} }{{x}^{\frac{\mathrm{2}}{\mathrm{3}}} }. \\$$
Commented by jagoll last updated on 03/Jan/20
$${sir}\:{i}'{m}\:{got}\:{f}^{−\mathrm{1}} \left({x}\right)=\:\frac{\left(\mathrm{4}{x}+\mathrm{2}\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{2}}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} \:−\mathrm{2}}{\mathrm{4}\left(\mathrm{4}{x}+\mathrm{2}\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{2}}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} } \\$$
Commented by MJS last updated on 03/Jan/20
$$\mathrm{how}? \\$$$${x}^{\mathrm{3}} +{px}+{q}=\mathrm{0} \\$$$$\mathrm{Cardano}'\mathrm{s}\:\mathrm{solution} \\$$$${x}={u}+{v} \\$$$${u}=\sqrt[{\mathrm{3}}]{−\frac{{q}}{\mathrm{2}}+\sqrt{\frac{{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{{q}^{\mathrm{2}} }{\mathrm{4}}}} \\$$$${v}=\sqrt[{\mathrm{3}}]{−\frac{{q}}{\mathrm{2}}−\sqrt{\frac{{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{{q}^{\mathrm{2}} }{\mathrm{4}}}} \\$$$${u},\:{v}\:\in\mathbb{R} \\$$$$\left(\mathrm{some}\:\mathrm{calculators}\:\mathrm{give}\:\mathrm{the}\:“\mathrm{wrong}''\:\mathrm{root}\right. \\$$$$\left.\mathrm{i}.\mathrm{e}.\:\sqrt[{\mathrm{3}}]{−\mathrm{8}}=−\mathrm{2}\:\mathrm{but}\:\sqrt[{\mathrm{3}}]{\mathrm{8e}^{\mathrm{i}\pi} }=\sqrt[{\mathrm{3}}]{\mathrm{8}}\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{3}}} =\mathrm{1}+\mathrm{i}\sqrt{\mathrm{3}}\right) \\$$