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Question Number 144282 by SOMEDAVONG last updated on 24/Jun/21

$$\mathrm{I}={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{{\sin }^{2021}\mathrm{x}}{{\sin }^{2021}\mathrm{x}+{\cos }^{2021}\mathrm{x}}\mathrm{dx}=?$$

Answered by som(math1967) last updated on 24/Jun/21

$$I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{{sin}^{2021}(\frac{\pi }{3}+\frac{\pi }{6}-x)}{{sin}^{2021}(\frac{\pi }{3}+\frac{\pi }{6}-x)+cos(\frac{\pi }{3}+\frac{\pi }{6}-x)}dx$$$$I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{{cos}^{2021}\mathrm{x}}{{\sin }^{2021}\mathrm{x}+{\cos }^{2021}\mathrm{x}}\mathrm{dx}$$$$\therefore 2I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{{cos}^{2021}\mathrm{x}+{sin}^{2021}x}{{\sin }^{2021}\mathrm{x}+{\cos }^{2021}\mathrm{x}}\mathrm{dx}$$$$2I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}dx$$$$I={\left[\frac{x}{2}\right]}_{\frac{\pi }{6}}^{\frac{\pi }{3}}=\frac{\pi }{12}ans$$

Commented by SOMEDAVONG last updated on 24/Jun/21

$$\mathrm{Thanks}\mathrm{sir}!$$

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