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

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(A) माना $I = \int_0^{\frac{\pi}{2}} \sqrt{	an x} \, dx$. गुणधर्म $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$ का उपयोग करने पर: $I = \int_0^{\frac{

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

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(A) माना $I = \int_0^{\frac{\pi}{2}} \sqrt{	an x} \, dx$. गुणधर्म $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$ का उपयोग करने पर: $I = \int_0^{\frac{\pi}{2}} \sqrt{	an(\frac{\pi}{2} - x)} \, dx = \int_0^{\frac{\pi}{2}} \sqrt{\cot x} \, dx$. $I$ के दोनों व्यंजकों को जोड़ने पर: $2I = \int_0^{\frac{\pi}{2}} (\sqrt{	an x} + \sqrt{\cot x}) \, dx = \int_0^{\frac{\pi}{2}} \frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} \, dx$. अंश और हर को $\sqrt{2}$ से गुणा करने पर: $2I = \sqrt{2} \int_0^{\frac{\pi}{2}} \frac{\sin x + \cos x}{\sqrt{2 \sin x \cos x}} \, dx = \sqrt{2} \int_0^{\frac{\pi}{2}} \frac{\sin x + \cos x}{\sqrt{1 - (\sin x - \cos x)^2}} \, dx$. माना $u = \sin x - \cos x$, तब $du = (\cos x + \sin x) \, dx$. जब $x=0, u=-1$; जब $x=\frac{\pi}{2}, u=1$. $2I = \sqrt{2} \int_{-1}^1 \frac{du}{\sqrt{1 - u^2}} = \sqrt{2} [\sin^{-1} u]_{-1}^1 = \sqrt{2} (\frac{\pi}{2} - (-\frac{\pi}{2})) = \sqrt{2} \pi$. अतः, $I = \frac{\sqrt{2} \pi}{2} = \frac{\pi}{\sqrt{2}}$.

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

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