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

Answered by Frix last updated on 04/Aug/24

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

Commented by Spillover last updated on 05/Aug/24

$$great$$

Answered by Spillover last updated on 05/Aug/24

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