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Question Number 76963 by peter frank last updated on 01/Jan/20

$$\int \cos 2\theta \ln \left(\frac{\cos \theta +\sin \theta }{\cos \theta -\sin \theta }\right)$$

Answered by MJS last updated on 02/Jan/20

$$\int \cos 2\theta \ln \frac{\cos \theta +\sin \theta }{\cos \theta -\sin \theta }d\theta =$$$$=\int \cos 2\theta \ln \frac{\cos 2\theta }{1-\sin 2\theta }d\theta =$$$$\mathrm{by}\mathrm{parts}$$$$u=\ln \frac{\cos 2\theta }{1-\sin 2\theta }\to u^{\prime }=\frac{2}{\cos 2\theta }$$$$v^{\prime }=\cos 2\theta \to v=\frac{\sin 2\theta }{2}$$$$=\frac{1}{2}\sin 2\theta \ln \frac{\cos 2\theta }{1-\sin 2\theta }-\int \tan 2\theta d\theta =$$$$=\frac{1}{2}\sin 2\theta \ln \frac{\cos 2\theta }{1-\sin 2\theta }-\frac{1}{2}\ln \cos 2\theta =$$$$=\frac{1}{2}(\sin 2\theta \ln \frac{\cos 2\theta }{1-\sin 2\theta }-\ln \cos 2\theta )+C$$

Commented by john santu last updated on 02/Jan/20

Commented by john santu last updated on 02/Jan/20

$$toMrMJS$$

Commented by peter frank last updated on 02/Jan/20

$$thankyou$$

Commented by MJS last updated on 02/Jan/20

$$\mathrm{I}\mathrm{get}\underset{x\to \mathrm{\infty }}{\lim }=\frac{1}{2}$$$$\mathrm{by}\mathrm{calculating}{f}^{-1}\left(x\right)\mathrm{and}\mathrm{then}\mathrm{approximating}$$

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

$$y=8x^3+3x$$$$\mathrm{exchanging}x\rightleftharpoons y$$$$y^3+\frac{3}{8}y-\frac{x}{8}=0$$$$D=\frac{p^3}{27}+\frac{q^2}{4}=\frac{1}{512}+\frac{x^2}{256}⩾0\mathrm{\forall }y\in \mathbb{R}\Rightarrow {\mathrm{Cardano}}^{\prime }\mathrm{s}\mathrm{firmula}$$$$\Rightarrow$$$$f^{-1}\left(x\right)=\frac{1}{2\sqrt[3]{4}}(\sqrt[3]{2x+\sqrt{4x^2+2}}-\sqrt[3]{-2x+\sqrt{4x^2+2}})$$$$g\left(x\right)=\frac{f^{-1}\left(8x\right)-f^{-1}\left(x\right)}{x^{1/3}}$$$$g\left({10}^1\right)=.526706...$$$$g\left({10}^2\right)=.505799...$$$$g\left({10}^3\right)=.501249...$$$$g\left({10}^4\right)=.500269...$$$$g\left({10}^5\right)=.500058...$$$$...$$

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

$$\mathrm{sorry}\mathrm{sir}.\mathrm{by}\mathrm{calculate}\mathrm{i}\mathrm{got}\mathrm{result}$$$$\frac{\sqrt[3]{4}}{2}\mathrm{sir}=\mathrm{not}\mathrm{same}\frac{1}{2}$$

Commented by mr W last updated on 02/Jan/20

$$tojohnsantusir:$$$$yourstepto$$$$\underset{x\to \mathrm{\infty }}{\lim }[\frac{8}{24{\left(g\left(x\right)\right)}^2+3}-\frac{1}{24{\left(h\left(x\right)\right)}^2\times 3}]\times 3x^{\frac{2}{3}}$$$$iscorrect.butthis{doesn}^{\prime }thelpyou$$$$further,sinceg\left(x\right)=f^{-1}\left(8x\right)and$$$$h\left(x\right)=f^{-1}\left(x\right)arestillunknown.$$$$yourlaststepto$$$$=3\times [\frac{8}{24}-\frac{1}{24}]$$$$iswrong.howcanyougetthisif$$$$you{don}^{\prime }texplictlyknowg\left(x\right)and$$$$h\left(x\right)?$$

Commented by john santu last updated on 02/Jan/20

$$oksir,iagree.ithoughtthedegrees$$$$g\left(x\right)andh\left(x\right)werethesame,but$$$$forgotthecoefficientsthatmight$$$$notbeequalto1.$$

Commented by mr W last updated on 02/Jan/20

$$thedegreesofg\left(x\right)andh\left(x\right)could$$$$bethesame,butyouhavehere$$$${\left(g\left(x\right)\right)}^{2}and{\left(h\left(x\right)\right)}^2,andwe{don}^{\prime }t$$$$evenknowthedegreesofthem.$$$$thereforewe{don}^{\prime }tknowthevaluesof$$$$\underset{x\to \mathrm{\infty }}{\lim }\frac{{\left(g\left(x\right)\right)}^2}{x^{\frac{2}{3}}}and\underset{x\to \mathrm{\infty }}{\lim }\frac{{\left(h\left(x\right)\right)}^2}{x^{\frac{2}{3}}}.$$

Commented by jagoll last updated on 03/Jan/20

$$sir{i}^{\prime }mgot{f}^{-1}\left(x\right)=\frac{{(4x+2\sqrt{4x^2+2})}^{\frac{2}{3}}-2}{4{(4x+2\sqrt{4x^2+2})}^{\frac{2}{3}}}$$

Commented by MJS last updated on 03/Jan/20

$$\mathrm{how}?$$$$x^3+px+q=0$$$${\mathrm{Cardano}}^{\prime }\mathrm{s}\mathrm{solution}$$$$x=u+v$$$$u=\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{p^3}{27}+\frac{q^2}{4}}}$$$$v=\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{p^3}{27}+\frac{q^2}{4}}}$$$$u,v\in \mathbb{R}$$$$(\mathrm{some}\mathrm{calculators}\mathrm{give}\mathrm{the}‘‘{\mathrm{wrong}}^{″}\mathrm{root}$$$$\mathrm{i}.\mathrm{e}.\sqrt[3]{-8}=-2\mathrm{but}\sqrt[3]{{8\mathrm{e}}^{\mathrm{i}\pi }}=\sqrt[3]{8}{\mathrm{e}}^{\mathrm{i}\frac{\pi }{3}}=1+\mathrm{i}\sqrt{3})$$

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