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

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(D) Let $I = \int_{-a}^{a} \sqrt{\frac{a - x}{a + x}} dx$. Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) dx$,where $a = -a$ and

Vedclass · Accepted Answer

$a \pi$

Vedclass · Answer

(D) Let $I = \int_{-a}^{a} \sqrt{\frac{a - x}{a + x}} dx$. Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) dx$,where $a = -a$ and $b = a$,we have $f(x) \rightarrow f(-a + a - x) = f(-x)$. $I = \int_{-a}^{a} \sqrt{\frac{a - (-x)}{a + (-x)}} dx = \int_{-a}^{a} \sqrt{\frac{a + x}{a - x}} dx$. Adding the two expressions for $I$: $2I = \int_{-a}^{a} \left( \sqrt{\frac{a - x}{a + x}} + \sqrt{\frac{a + x}{a - x}} \right) dx$. $2I = \int_{-a}^{a} \frac{(a - x) + (a + x)}{\sqrt{(a + x)(a - x)}} dx = \int_{-a}^{a} \frac{2a}{\sqrt{a^2 - x^2}} dx$. Since the integrand is an even function,$2I = 2 \int_{0}^{a} \frac{2a}{\sqrt{a^2 - x^2}} dx$. $I = 2a \int_{0}^{a} \frac{1}{\sqrt{a^2 - x^2}} dx$. $I = 2a [\sin^{-1}(\frac{x}{a})]_{0}^{a}$. $I = 2a (\sin^{-1}(1) - \sin^{-1}(0)) = 2a (\frac{\pi}{2} - 0) = a \pi$.

$\int_{-a}^{a} \sqrt{\frac{a - x}{a + x}} dx =$

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