If int_{ - a}^a {sqrt {frac{{a - x}}{{a + x}} | vedclass

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(D) Let $I = \int_{ - a}^a {\sqrt {\frac{{a - x}}{{a + x}}} dx} $. To evaluate this,we substitute $x = a \cos 	heta$. Then $dx = -a \sin 	heta \, d\

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$a$

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(D) Let $I = \int_{ - a}^a {\sqrt {\frac{{a - x}}{{a + x}}} dx} $. To evaluate this,we substitute $x = a \cos 	heta$. Then $dx = -a \sin 	heta \, d	heta$. When $x = -a$,$\cos 	heta = -1$,so $	heta = \pi$. When $x = a$,$\cos 	heta = 1$,so $	heta = 0$. $I = \int_{\pi}^0 \sqrt{\frac{a - a \cos 	heta}{a + a \cos 	heta}} (-a \sin 	heta) \, d	heta$. $I = \int_0^{\pi} \sqrt{\frac{1 - \cos 	heta}{1 + \cos 	heta}} (a \sin 	heta) \, d	heta$. Using half-angle formulas,$\sqrt{\frac{1 - \cos 	heta}{1 + \cos 	heta}} = \sqrt{\frac{2 \sin^2(	heta/2)}{2 \cos^2(	heta/2)}} = 	an(	heta/2)$. $I = a \int_0^{\pi} 	an(	heta/2) \cdot 2 \sin(	heta/2) \cos(	heta/2) \, d	heta$. $I = 2a \int_0^{\pi} \frac{\sin(	heta/2)}{\cos(	heta/2)} \cdot \sin(	heta/2) \cos(	heta/2) \, d	heta$. $I = 2a \int_0^{\pi} \sin^2(	heta/2) \, d	heta$. Using $\sin^2(	heta/2) = \frac{1 - \cos 	heta}{2}$,we get $I = 2a \int_0^{\pi} \frac{1 - \cos 	heta}{2} \, d	heta = a [	heta - \sin 	heta]_0^{\pi} = a(\pi - 0) = a\pi$. Comparing $a\pi$ with $k\pi$,we find $k = a$.

If $\int_{ - a}^a {\sqrt {\frac{{a - x}}{{a + x}}} \,dx = k\pi ,} $ then $k = $

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