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

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(C) ધારો કે $I = \int_0^{\pi /2} \frac{\sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx$ .....$(i)$ગુણધર્મ $\int_0^a f(x) \, dx = \int_0^a f(a-x) \

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(C) ધારો કે $I = \int_0^{\pi /2} \frac{\sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx$ .....$(i)$ગુણધર્મ $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$ નો ઉપયોગ કરતા:$I = \int_0^{\pi /2} \frac{\sqrt{\cos(\pi/2 - x)}}{\sqrt{\sin(\pi/2 - x)} + \sqrt{\cos(\pi/2 - x)}} \, dx$કારણ કે $\cos(\pi/2 - x) = \sin x$ અને $\sin(\pi/2 - x) = \cos x$ હોવાથી:$I = \int_0^{\pi /2} \frac{\sqrt{\sin x}}{\sqrt{\cos x} + \sqrt{\sin x}} \, dx$ .....$(ii)$સમીકરણ $(i)$ અને $(ii)$ નો સરવાળો કરતા:$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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