If int_{0}^{frac{pi}{3}} frac{tan theta}{sqrt | vedclass

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(A) Let $I = \int_{0}^{\frac{\pi}{3}} \frac{	an 	heta}{\sqrt{2 k \sec 	heta}} d 	heta$. $= \frac{1}{\sqrt{2 k}} \int_{0}^{\frac{\pi}{3}} \frac{\si

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$2$

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(A) Let $I = \int_{0}^{\frac{\pi}{3}} \frac{	an 	heta}{\sqrt{2 k \sec 	heta}} d 	heta$. $= \frac{1}{\sqrt{2 k}} \int_{0}^{\frac{\pi}{3}} \frac{\sin 	heta}{\cos 	heta} 	imes \sqrt{\cos 	heta} d 	heta$. $= \frac{1}{\sqrt{2 k}} \int_{0}^{\frac{\pi}{3}} \frac{\sin 	heta}{\sqrt{\cos 	heta}} d 	heta$. Let $\cos 	heta = t$,then $-\sin 	heta d 	heta = dt$,so $\sin 	heta d 	heta = -dt$. When $	heta = 0$,$t = 1$. When $	heta = \frac{\pi}{3}$,$t = \frac{1}{2}$. $I = \frac{-1}{\sqrt{2 k}} \int_{1}^{\frac{1}{2}} t^{-\frac{1}{2}} dt = \frac{1}{\sqrt{2 k}} \int_{\frac{1}{2}}^{1} t^{-\frac{1}{2}} dt$. $I = \frac{1}{\sqrt{2 k}} [2\sqrt{t}]_{\frac{1}{2}}^{1} = \frac{2}{\sqrt{2 k}} (1 - \frac{1}{\sqrt{2}}) = \frac{\sqrt{2}}{\sqrt{k}} (1 - \frac{1}{\sqrt{2}})$. Given $I = 1 - \frac{1}{\sqrt{2}}$,we have $\frac{\sqrt{2}}{\sqrt{k}} = 1$,which implies $\sqrt{k} = \sqrt{2}$,so $k = 2$.

If $\int_{0}^{\frac{\pi}{3}} \frac{\tan \theta}{\sqrt{2 k \sec \theta}} d \theta = 1 - \frac{1}{\sqrt{2}}$,$(k > 0)$,then the value of $k$ is

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