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

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Question

(A) Let $x = \cos^2 	heta$. Then $dx = -2 \cos 	heta \sin 	heta \, d	heta$. Substituting these into the integral: $\int \sqrt{\frac{1 - \cos 	het

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $x = \cos^2 	heta$. Then $dx = -2 \cos 	heta \sin 	heta \, d	heta$. Substituting these into the integral: $\int \sqrt{\frac{1 - \cos 	heta}{1 + \cos 	heta}} (-2 \cos 	heta \sin 	heta) \, d	heta = \int \sqrt{\frac{2 \sin^2(	heta/2)}{2 \cos^2(	heta/2)}} (-2 \cos 	heta \sin 	heta) \, d	heta$ $= \int 	an(	heta/2) (-2 \cos 	heta \sin 	heta) \, d	heta = \int \frac{\sin(	heta/2)}{\cos(	heta/2)} (-4 \sin(	heta/2) \cos(	heta/2) \cos 	heta) \, d	heta$ $= -4 \int \sin^2(	heta/2) \cos 	heta \, d	heta = -2 \int (1 - \cos 	heta) \cos 	heta \, d	heta$ $= -2 \int (\cos 	heta - \cos^2 	heta) \, d	heta = -2 \int \cos 	heta \, d	heta + \int (1 + \cos 2	heta) \, d	heta$ $= -2 \sin 	heta + 	heta + \frac{1}{2} \sin 2	heta + c = 	heta + \sin 	heta \cos 	heta - 2 \sin 	heta + c$ Since $x = \cos^2 	heta$,we have $\cos 	heta = \sqrt{x}$ and $\sin 	heta = \sqrt{1 - x}$. Substituting back: $\cos^{-1}\sqrt{x} + \sqrt{1 - x} \cdot \sqrt{x} - 2 \sqrt{1 - x} + c = \cos^{-1}\sqrt{x} + \sqrt{1 - x}(\sqrt{x} - 2) + c$.

$\int \sqrt{\frac{1 - \sqrt{x}}{1 + \sqrt{x}}} \, dx = $

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