The value of the definite integral intlimits_ | vedclass

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(B) Let $I = \int\limits_0^{\frac{\pi }{2}} {\sqrt {	an x} \,dx} \quad ....(1)$Using the property $\int\limits_0^a f(x) dx = \int\limits_0^a f(a-x) d

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

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(B) Let $I = \int\limits_0^{\frac{\pi }{2}} {\sqrt {	an x} \,dx} \quad ....(1)$Using the property $\int\limits_0^a f(x) dx = \int\limits_0^a f(a-x) dx$,we get:$I = \int\limits_0^{\frac{\pi }{2}} {\sqrt {	an(\frac{\pi}{2} - x)} \,dx} = \int\limits_0^{\frac{\pi }{2}} {\sqrt {\cot x} \,dx} \quad ....(2)$Adding $(1)$ and $(2)$:$2I = \int\limits_0^{\frac{\pi }{2}} {(\sqrt {	an x} + \sqrt {\cot x}) \,dx} = \int\limits_0^{\frac{\pi }{2}} {\frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} \,dx}$Multiplying numerator and denominator by $\sqrt{2}$:$2I = \sqrt{2} \int\limits_0^{\frac{\pi }{2}} {\frac{\sin x + \cos x}{\sqrt{2 \sin x \cos x}} \,dx} = \sqrt{2} \int\limits_0^{\frac{\pi }{2}} {\frac{\sin x + \cos x}{\sqrt{1 - (\sin x - \cos x)^2}} \,dx}$Let $t = \sin x - \cos x$,then $dt = (\cos x + \sin x) dx$. When $x=0, t=-1$; when $x=\frac{\pi}{2}, t=1$.$2I = \sqrt{2} \int\limits_{-1}^1 {\frac{dt}{\sqrt{1 - t^2}}} = \sqrt{2} [\sin^{-1} t]_{-1}^1 = \sqrt{2} (\frac{\pi}{2} - (-\frac{\pi}{2})) = \sqrt{2} \pi$$I = \frac{\sqrt{2} \pi}{2} = \frac{\pi}{\sqrt{2}}$.

The value of the definite integral $\int\limits_0^{\frac{\pi }{2}} {\sqrt {\tan x} \,dx} $ is

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