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

Solution

$I = \int_{- a}^{a} \sqrt{\frac{a - x}{a + x}} dx = \int_{- a}^{a} \frac{a - x}{\sqrt{a^{2} - x^{2}}} dx = \int_{- a}^{a} \frac{a}{\sqrt{a^{2} - x^{2}}} dx - \int_{- a}^{a} \frac{x}{\sqrt{a^{2} - x^{2}}} dx = I_{1} + I_{2} Where I_{1} = \int_{- a}^{a} \frac{a}{\sqrt{a^{2} - x^{2}}} dx, which is the integral of an even function And I_{2} = \int_{- a}^{a} \frac{x}{\sqrt{a^{2} - x^{2}}} dx, which is the integral of an even function, and so I_{2} = 0$

Now,

$I = I_{1} = \int_{- a}^{a} \frac{a}{\sqrt{a^{2} - x^{2}}} dx = 2 \int_{0}^{a} \frac{a}{\sqrt{a^{2} - x^{2}}} dx = 2 a \int_{0}^{a} \frac{1}{\sqrt{a^{2} - x^{2}}} dx = 2 a . {[\sin^{- 1} (\frac{x}{a})]}_{0}^{a} = 2 a . [\sin^{- 1} 1 - \sin^{- 1} 0] = 2 a (\frac{π}{2}) = πa$

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