int_{0}^{frac{1}{2}} frac{dx}{(1+x^{2}) sqrt | vedclass

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(A) Let $ I = \int_{0}^{1/2} \frac{dx}{(1+x^{2}) \sqrt{1-x^{2}}} $.Substitute $ x = \sin 	heta $,then $ dx = \cos 	heta d	heta $.When $ x = 0, 	he

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$ \frac{1}{\sqrt{2}} 	an^{-1} \sqrt{\frac{2}{3}} $

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(A) Let $ I = \int_{0}^{1/2} \frac{dx}{(1+x^{2}) \sqrt{1-x^{2}}} $.Substitute $ x = \sin 	heta $,then $ dx = \cos 	heta d	heta $.When $ x = 0, 	heta = 0 $. When $ x = 1/2, 	heta = \pi/6 $.The integral becomes $ I = \int_{0}^{\pi/6} \frac{\cos 	heta d	heta}{(1+\sin^{2} 	heta) \sqrt{1-\sin^{2} 	heta}} = \int_{0}^{\pi/6} \frac{\cos 	heta d	heta}{(1+\sin^{2} 	heta) \cos 	heta} = \int_{0}^{\pi/6} \frac{d	heta}{1+\sin^{2} 	heta} $.Divide numerator and denominator by $ \cos^{2} 	heta $: $ I = \int_{0}^{\pi/6} \frac{\sec^{2} 	heta d	heta}{\sec^{2} 	heta + 	an^{2} 	heta} = \int_{0}^{\pi/6} \frac{\sec^{2} 	heta d	heta}{1 + 2	an^{2} 	heta} $.Let $ u = \sqrt{2} 	an 	heta $,then $ du = \sqrt{2} \sec^{2} 	heta d	heta $,so $ \sec^{2} 	heta d	heta = \frac{du}{\sqrt{2}} $.Limits: $ 	heta = 0 \implies u = 0 $; $ 	heta = \pi/6 \implies u = \sqrt{2} 	an(\pi/6) = \sqrt{2} \cdot \frac{1}{\sqrt{3}} = \sqrt{\frac{2}{3}} $.Thus,$ I = \int_{0}^{\sqrt{2/3}} \frac{1}{1+u^{2}} \cdot \frac{du}{\sqrt{2}} = \frac{1}{\sqrt{2}} [	an^{-1} u]_{0}^{\sqrt{2/3}} = \frac{1}{\sqrt{2}} 	an^{-1} \sqrt{\frac{2}{3}} $.

$ \int_{0}^{\frac{1}{2}} \frac{dx}{(1+x^{2}) \sqrt{1-x^{2}}} $ is equal to

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