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Question Number 186637 by Mingma last updated on 07/Feb/23

Answered by mr W last updated on 07/Feb/23

$$={\int }_0^{\frac{\pi }{4}}\frac{dx}{2-(1-2{\sin }^{2}x)}$$$$={\int }_0^{\frac{\pi }{4}}\frac{dx}{2-\cos 2x}$$$$=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\frac{dx}{2-\cos x}$$$$=\frac{1}{\sqrt{3}}{\left[{\tan }^{-1}\left(\sqrt{3}\tan \frac{x}{2}\right)\right]}_0^{\frac{\pi }{2}}$$$$=\frac{\pi }{3\sqrt{3}}$$

Answered by ARUNG_Brandon_MBU last updated on 08/Feb/23

$$I={\int }_0^{\frac{\pi }{4}}\frac{dx}{1+{2\sin }^{2}x}={\int }_0^{\frac{\pi }{4}}\frac{{\sec }^{2}x}{{\sec }^{2}x+{2\tan }^{2}x}dx$$$$={\int }_0^{\frac{\pi }{4}}\frac{{\sec }^{2}x}{{3\tan }^{2}x+1}dx={\int }_0^1\frac{dt}{3t^2+1}$$$$=\frac{1}{\sqrt{3}}{\left[\arctan \left(\sqrt{3}t\right)\right]}_0^1=\frac{\pi }{3\sqrt{3}}$$

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