int sqrt{frac{1-sqrt{x}}{1+sqrt{x}}} , dx = | vedclass

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Question

(B) Let $I = \int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} \, dx$. Multiply numerator and denominator by $\sqrt{1-\sqrt{x}}$: $I = \int \sqrt{\frac{(1-\sq

Vedclass · Accepted Answer

$-2 \sqrt{1-x} + \cos^{-1} \sqrt{x} + \sqrt{x(1-x)} + C$

Vedclass · Answer

(B) Let $I = \int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} \, dx$. Multiply numerator and denominator by $\sqrt{1-\sqrt{x}}$: $I = \int \sqrt{\frac{(1-\sqrt{x})^2}{1-x}} \, dx = \int \frac{1-\sqrt{x}}{\sqrt{1-x}} \, dx$. Split the integral: $I = \int \frac{1}{\sqrt{1-x}} \, dx - \int \frac{\sqrt{x}}{\sqrt{1-x}} \, dx$. The first part is $-2\sqrt{1-x}$. For the second part,let $x = \cos^2 	heta$,so $dx = -2 \cos 	heta \sin 	heta \, d	heta$. $\int \frac{\sqrt{x}}{\sqrt{1-x}} \, dx = \int \frac{\cos 	heta}{\sin 	heta} (-2 \cos 	heta \sin 	heta) \, d	heta = -2 \int \cos^2 	heta \, d	heta = -\int (1 + \cos 2	heta) \, d	heta = -(	heta + \frac{\sin 2	heta}{2}) = -	heta - \sin 	heta \cos 	heta$. Substituting back: $I = -2\sqrt{1-x} - (-	heta - \sin 	heta \cos 	heta) + C = -2\sqrt{1-x} + 	heta + \sin 	heta \cos 	heta + C$. Since $x = \cos^2 	heta$,$	heta = \cos^{-1} \sqrt{x}$,$\cos 	heta = \sqrt{x}$,and $\sin 	heta = \sqrt{1-x}$. Thus,$I = -2\sqrt{1-x} + \cos^{-1} \sqrt{x} + \sqrt{x(1-x)} + C$.

$\int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} \, dx = $ (where $C$ is a constant of integration)

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