Question-210263

Question Number 210263 by Spillover last updated on 04/Aug/24

Answered by Frix last updated on 04/Aug/24

With t=tan (x^3 /2) and α=ln 12 we get −(4/3)∫(t/((t^2 +1)(t^2 sin α −2t(1−cos α)−sin α)))dt= Let u=cos α ∧ v=sin α =(v/(3(1−u)))∫(t/(t^2 +1))dt+(1/3)∫(dt/(t^2 +1))− −((u+1)/6)∫(dt/(vt+u−1−(√(2(1−u)))))− −((u+1)/6)∫(dt/(vt+u−1+(√(2(1−u))))) which are easy to solve. I finally get ((ln 2)/3)−((ln 3)/6)

$$\mathrm{With}\:{t}=\mathrm{tan}\:\frac{{x}^{\mathrm{3}} }{\mathrm{2}}\:\mathrm{and}\:\alpha=\mathrm{ln}\:\mathrm{12}\:\mathrm{we}\:\mathrm{get} \\ $$$$−\frac{\mathrm{4}}{\mathrm{3}}\int\frac{{t}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)\left({t}^{\mathrm{2}} \mathrm{sin}\:\alpha\:−\mathrm{2}{t}\left(\mathrm{1}−\mathrm{cos}\:\alpha\right)−\mathrm{sin}\:\alpha\right)}{dt}= \\ $$$$\mathrm{Let}\:{u}=\mathrm{cos}\:\alpha\:\wedge\:{v}=\mathrm{sin}\:\alpha \\ $$$$=\frac{{v}}{\mathrm{3}\left(\mathrm{1}−{u}\right)}\int\frac{{t}}{{t}^{\mathrm{2}} +\mathrm{1}}{dt}+\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}− \\ $$$$−\frac{{u}+\mathrm{1}}{\mathrm{6}}\int\frac{{dt}}{{vt}+{u}−\mathrm{1}−\sqrt{\mathrm{2}\left(\mathrm{1}−{u}\right)}}− \\ $$$$−\frac{{u}+\mathrm{1}}{\mathrm{6}}\int\frac{{dt}}{{vt}+{u}−\mathrm{1}+\sqrt{\mathrm{2}\left(\mathrm{1}−{u}\right)}} \\ $$$$\mathrm{which}\:\mathrm{are}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{solve}. \\ $$$$\mathrm{I}\:\mathrm{finally}\:\mathrm{get} \\ $$$$\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{3}}−\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{6}} \\ $$

Commented by Spillover last updated on 05/Aug/24

$${great} \\ $$

Answered by Spillover last updated on 05/Aug/24

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Question-210263

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