int_{1 / 2}^{1 / sqrt{2}} frac{1}{left(x+sqrt | vedclass

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

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$\log (3-\sqrt{3})$

Vedclass · Answer

(D) Let $I = \int_{1/2}^{1/\sqrt{2}} \frac{1}{(x+\sqrt{1-x^2})(1-x^2)} dx$. Substitute $x = \sin 	heta$,then $dx = \cos 	heta d	heta$. When $x = 1/2$,$	heta = \pi/6$. When $x = 1/\sqrt{2}$,$	heta = \pi/4$. $I = \int_{\pi/6}^{\pi/4} \frac{\cos 	heta d	heta}{(\sin 	heta + \cos 	heta) \cos^2 	heta} = \int_{\pi/6}^{\pi/4} \frac{d	heta}{(\sin 	heta + \cos 	heta) \cos 	heta}$. Divide numerator and denominator by $\cos 	heta$: $I = \int_{\pi/6}^{\pi/4} \frac{\sec^2 	heta d	heta}{	an 	heta + 1}$. Let $u = 	an 	heta$,then $du = \sec^2 	heta d	heta$. When $	heta = \pi/6$,$u = 1/\sqrt{3}$. When $	heta = \pi/4$,$u = 1$. $I = \int_{1/\sqrt{3}}^{1} \frac{du}{u+1} = [\log |u+1|]_{1/\sqrt{3}}^{1} = \log(2) - \log(1 + 1/\sqrt{3}) = \log \left( \frac{2}{1 + 1/\sqrt{3}} ight) = \log \left( \frac{2\sqrt{3}}{\sqrt{3}+1} ight)$. Rationalizing the denominator: $\frac{2\sqrt{3}(\sqrt{3}-1)}{3-1} = \sqrt{3}(\sqrt{3}-1) = 3-\sqrt{3}$. Thus,$I = \log(3-\sqrt{3})$.

$\int_{1 / 2}^{1 / \sqrt{2}} \frac{1}{\left(x+\sqrt{1-x^2}\right)\left(1-x^2\right)} d x=$

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