The value of int_{0}^{1} frac{dx}{x + sqrt{1 | vedclass

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(D) Let $I = \int_{0}^{1} \frac{dx}{x + \sqrt{1 - x^2}}$.Substitute $x = \sin 	heta$,then $dx = \cos 	heta \, d	heta$.When $x = 0$,$	heta = 0$. Wh

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$\frac{\pi}{4}$

Vedclass · Answer

(D) Let $I = \int_{0}^{1} \frac{dx}{x + \sqrt{1 - x^2}}$.Substitute $x = \sin 	heta$,then $dx = \cos 	heta \, d	heta$.When $x = 0$,$	heta = 0$. When $x = 1$,$	heta = \frac{\pi}{2}$.$I = \int_{0}^{\pi/2} \frac{\cos 	heta \, d	heta}{\sin 	heta + \cos 	heta}$.Using the property $\int_{0}^{a} f(	heta) \, d	heta = \int_{0}^{a} f(a - 	heta) \, d	heta$,we get:$I = \int_{0}^{\pi/2} \frac{\cos(\pi/2 - 	heta) \, d	heta}{\sin(\pi/2 - 	heta) + \cos(\pi/2 - 	heta)} = \int_{0}^{\pi/2} \frac{\sin 	heta \, d	heta}{\cos 	heta + \sin 	heta}$.Adding the two expressions for $I$:$2I = \int_{0}^{\pi/2} \frac{\cos 	heta + \sin 	heta}{\sin 	heta + \cos 	heta} \, d	heta = \int_{0}^{\pi/2} 1 \, d	heta = [	heta]_{0}^{\pi/2} = \frac{\pi}{2}$.Therefore,$I = \frac{\pi}{4}$.

The value of $\int_{0}^{1} \frac{dx}{x + \sqrt{1 - x^2}}$ is

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