The value of the definite integral int_{0}^{p | vedclass

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(B) Let $I = \int_{0}^{\pi/2} \sqrt{	an x} \, dx$. Using the property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$,we get: $I = \int_{0}^{\p

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$\frac{\pi}{\sqrt{2}}$

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(B) Let $I = \int_{0}^{\pi/2} \sqrt{	an x} \, dx$. Using the property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$,we get: $I = \int_{0}^{\pi/2} \sqrt{	an(\pi/2 - x)} \, dx = \int_{0}^{\pi/2} \sqrt{\cot x} \, dx$. Adding the two expressions for $I$: $2I = \int_{0}^{\pi/2} (\sqrt{	an x} + \sqrt{\cot x}) \, dx = \int_{0}^{\pi/2} \frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} \, dx$. Multiply numerator and denominator by $\sqrt{2}$: $2I = \sqrt{2} \int_{0}^{\pi/2} \frac{\sin x + \cos x}{\sqrt{2 \sin x \cos x}} \, dx = \sqrt{2} \int_{0}^{\pi/2} \frac{\sin x + \cos x}{\sqrt{1 - (\sin x - \cos x)^2}} \, dx$. Let $u = \sin x - \cos x$,then $du = (\cos x + \sin x) \, dx$. When $x=0, u=-1$; when $x=\pi/2, u=1$. $2I = \sqrt{2} \int_{-1}^{1} \frac{du}{\sqrt{1-u^2}} = \sqrt{2} [\sin^{-1}(u)]_{-1}^{1} = \sqrt{2} (\frac{\pi}{2} - (-\frac{\pi}{2})) = \sqrt{2} \pi$. Therefore,$I = \frac{\sqrt{2} \pi}{2} = \frac{\pi}{\sqrt{2}}$.

The value of the definite integral $\int_{0}^{\pi/2} \sqrt{\tan x} \, dx$ is:

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