int_0^{frac{pi}{2}} sqrt{tan x} , dx = | vedclass

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(A) Let $I = \int_0^{\frac{\pi}{2}} \sqrt{	an x} \, dx$. Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$, we get: $I = \int_0^{\frac

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

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(A) Let $I = \int_0^{\frac{\pi}{2}} \sqrt{	an x} \, dx$. Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$, we get: $I = \int_0^{\frac{\pi}{2}} \sqrt{	an(\frac{\pi}{2} - x)} \, dx = \int_0^{\frac{\pi}{2}} \sqrt{\cot x} \, dx$. Adding the two expressions for $I$: $2I = \int_0^{\frac{\pi}{2}} (\sqrt{	an x} + \sqrt{\cot x}) \, dx = \int_0^{\frac{\pi}{2}} \frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} \, dx$. Multiply numerator and denominator by $\sqrt{2}$: $2I = \sqrt{2} \int_0^{\frac{\pi}{2}} \frac{\sin x + \cos x}{\sqrt{2 \sin x \cos x}} \, dx = \sqrt{2} \int_0^{\frac{\pi}{2}} \frac{\sin x + \cos x}{\sqrt{1 - (\sin x - \cos x)^2}} \, dx$. Let $u = \sin x - \cos x$, then $du = (\cos x + \sin x) \, dx$. When $x=0, u=-1$; when $x=\frac{\pi}{2}, u=1$. $2I = \sqrt{2} \int_{-1}^1 \frac{du}{\sqrt{1 - u^2}} = \sqrt{2} [\sin^{-1} u]_{-1}^1 = \sqrt{2} (\frac{\pi}{2} - (-\frac{\pi}{2})) = \sqrt{2} \pi$. Thus, $I = \frac{\sqrt{2} \pi}{2} = \frac{\pi}{\sqrt{2}}$.

$\int_0^{\frac{\pi}{2}} \sqrt{\tan x} \, dx =$

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