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Question Number 156793 by alcohol last updated on 15/Oct/21

$${\int }_0^{\frac{\pi }{2}}\frac{sin2x}{2-{sin}^{2}2x}dx$$

Answered by FongXD last updated on 15/Oct/21

$$={\int }_0^{\frac{\pi }{2}}\frac{\sin 2\mathrm{x}}{1+{\cos }^{2}2\mathrm{x}}\mathrm{dx}=-\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\frac{{\left(\cos 2\mathrm{x}\right)}^{\prime }}{1+{\cos }^{2}2\mathrm{x}}\mathrm{dx}$$$$=-\frac{1}{2}{\left[\arctan \left(\cos 2\mathrm{x}\right)\right]}_0^{\frac{\pi }{2}}=-\frac{1}{2}(-\frac{\pi }{4}-\frac{\pi }{4})=\frac{\pi }{4}$$$$\mathrm{therefore}.{\int }_0^{\frac{\pi }{2}}\frac{\sin 2\mathrm{x}}{2-{\sin }^{2}2\mathrm{x}}\mathrm{dx}=\frac{\pi }{4}$$

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