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Question Number 137359 by rexford last updated on 01/Apr/21

Answered by Ar Brandon last updated on 01/Apr/21

$$\mathcal{I}={\int }_{\sqrt[3]{\log 3}}^{\sqrt[3]{\log 4}}\frac{{\mathrm{x}}^2{\mathrm{sinx}}^3}{{\mathrm{sinx}}^3+\sin (\log 12-{\mathrm{x}}^3)}\mathrm{dx}$$$$\stackrel{(\mathrm{u}={\mathrm{x}}^3)}{=}\frac{1}{3}{\int }_{\log 3}^{\log 4}\frac{\sin \left(\mathrm{u}\right)}{\sin \left(\mathrm{u}\right)+\sin (\log 12-\mathrm{u})}\mathrm{du}$$$$=\frac{1}{3}{\int }_{\log 3}^{\log 4}\frac{\sin (1\mathrm{og}12-\mathrm{u})}{\sin \left(\mathrm{u}\right)+\sin (\log 12-\mathrm{u})}\mathrm{du}$$$$2\mathcal{I}=\frac{1}{3}{\int }_{\log 3}^{\log 4}\frac{\sin \left(\mathrm{u}\right)+\sin (\log 12-\mathrm{u})}{\sin \left(\mathrm{u}\right)+\sin (\log 12-\mathrm{u})}\mathrm{du}$$$$=\frac{1}{3}\log \left(\frac{4}{3}\right)\Rightarrow \mathcal{I}=\frac{1}{6}\log \left(\frac{4}{3}\right)$$

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