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

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

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$\frac{\pi}{4}$

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

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

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