int_{pi /6}^{pi /3} frac{dx}{1 + sqrt{tan x}} | vedclass

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Question

(A) ધારો કે $I = \int_{\pi /6}^{\pi /3} \frac{dx}{1 + \sqrt{	an x}}$.$I = \int_{\pi /6}^{\pi /3} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}}

Vedclass · Accepted Answer

$\pi /12$

Vedclass · Answer

(A) ધારો કે $I = \int_{\pi /6}^{\pi /3} \frac{dx}{1 + \sqrt{	an x}}$.$I = \int_{\pi /6}^{\pi /3} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx$ ...$(i)$ગુણધર્મ $\int_a^b f(x) dx = \int_a^b f(a + b - x) dx$ નો ઉપયોગ કરતા,આપણને મળે છે:$I = \int_{\pi /6}^{\pi /3} \frac{\sqrt{\cos(\pi/2 - x)}}{\sqrt{\cos(\pi/2 - x)} + \sqrt{\sin(\pi/2 - x)}} dx$$I = \int_{\pi /6}^{\pi /3} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$ ...(ii)$(i)$ અને $(ii)$ નો સરવાળો કરતા:$2I = \int_{\pi /6}^{\pi /3} \frac{\sqrt{\cos x} + \sqrt{\sin x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx$$2I = \int_{\pi /6}^{\pi /3} 1 dx$$2I = [x]_{\pi /6}^{\pi /3} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$$I = \frac{\pi}{12}$.

$\int_{\pi /6}^{\pi /3} \frac{dx}{1 + \sqrt{\tan x}} = $

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