int_0^2 sqrt{frac{2 + x}{2 - x}} ,dx = | vedclass

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Question

(A) Let $I = \int_0^2 \sqrt{\frac{2 + x}{2 - x}} \,dx$. Put $x = 2 \cos 	heta$,so $dx = -2 \sin 	heta \,d	heta$. When $x = 0$,$\cos 	heta = 0 \Rig

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$\pi + 2$

Vedclass · Answer

(A) Let $I = \int_0^2 \sqrt{\frac{2 + x}{2 - x}} \,dx$. Put $x = 2 \cos 	heta$,so $dx = -2 \sin 	heta \,d	heta$. When $x = 0$,$\cos 	heta = 0 \Rightarrow 	heta = \frac{\pi}{2}$. When $x = 2$,$\cos 	heta = 1 \Rightarrow 	heta = 0$. Substituting these into the integral: $I = \int_{\pi/2}^0 \sqrt{\frac{2 + 2 \cos 	heta}{2 - 2 \cos 	heta}} (-2 \sin 	heta) \,d	heta$ $I = 2 \int_0^{\pi/2} \sqrt{\frac{1 + \cos 	heta}{1 - \cos 	heta}} \sin 	heta \,d	heta$ Using half-angle identities $1 + \cos 	heta = 2 \cos^2(	heta/2)$ and $1 - \cos 	heta = 2 \sin^2(	heta/2)$: $I = 2 \int_0^{\pi/2} \frac{\cos(	heta/2)}{\sin(	heta/2)} \cdot 2 \sin(	heta/2) \cos(	heta/2) \,d	heta$ $I = 4 \int_0^{\pi/2} \cos^2(	heta/2) \,d	heta$ Using $\cos^2(	heta/2) = \frac{1 + \cos 	heta}{2}$: $I = 4 \int_0^{\pi/2} \frac{1 + \cos 	heta}{2} \,d	heta = 2 \int_0^{\pi/2} (1 + \cos 	heta) \,d	heta$ $I = 2 [	heta + \sin 	heta]_0^{\pi/2} = 2 [(\frac{\pi}{2} + 1) - (0 + 0)] = \pi + 2$.

$\int_0^2 \sqrt{\frac{2 + x}{2 - x}} \,dx = $

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