int_0^{pi /2} frac{sqrt{cot x}}{sqrt{cot x} + | vedclass

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(C) Let $I = \int_0^{\pi /2} \frac{\sqrt{\cot x}}{\sqrt{\cot x} + \sqrt{	an x}} \, dx$ ..... $(i)$Using the property $\int_0^a f(x) \, dx = \int_0^a

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$\frac{\pi}{4}$

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(C) Let $I = \int_0^{\pi /2} \frac{\sqrt{\cot x}}{\sqrt{\cot x} + \sqrt{	an x}} \, dx$ ..... $(i)$Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we get:$I = \int_0^{\pi /2} \frac{\sqrt{\cot(\pi/2 - x)}}{\sqrt{\cot(\pi/2 - x)} + \sqrt{	an(\pi/2 - x)}} \, dx$Since $\cot(\pi/2 - x) = 	an x$ and $	an(\pi/2 - x) = \cot x$,we have:$I = \int_0^{\pi /2} \frac{\sqrt{	an x}}{\sqrt{	an x} + \sqrt{\cot x}} \, dx$ ..... $(ii)$Adding $(i)$ and $(ii)$:$2I = \int_0^{\pi /2} \frac{\sqrt{\cot x} + \sqrt{	an x}}{\sqrt{\cot x} + \sqrt{	an x}} \, dx$$2I = \int_0^{\pi /2} 1 \, dx$$2I = [x]_0^{\pi /2} = \frac{\pi}{2} - 0 = \frac{\pi}{2}$$I = \frac{\pi}{4}$

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

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