∫−aa a−xa+xdx is equal to

I=∫−aa a−xa+xdx=∫−aa a−xa2−x2dx=a∫−aa dxa2−x2−∫−aa xdxa2−x2=a⋅2∫0a dxa2−x2−0 ∵xa2−x2 is an odd function =2asin−1⁡xa0a⇒2asin−1⁡(1)−sin−1⁡(0)=πa

∫−aa a−xa+xdx is equal to

I=∫−aa a−xa+xdx=∫−aa a−xa2−x2dx=a∫−aa dxa2−x2−∫−aa xdxa2−x2=a⋅2∫0a dxa2−x2−0 ∵xa2−x2 is an odd function =2asin−1⁡xa0a⇒2asin−1⁡(1)−sin−1⁡(0)=πa

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$\int_{- a}^{a} \sqrt{\frac{a - x}{a + x}} dx$ is equal to

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$π$

$\frac{aπ}{2}$

$aπ$

answer is D.

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Detailed Solution

$\begin{array}{l} I = \int_{- a}^{a} \sqrt{\frac{a - x}{a + x}} dx = \int_{- a}^{a} \frac{a - x}{\sqrt{a^{2} - x^{2}}} dx \\ = a \int_{- a}^{a} \frac{dx}{\sqrt{a^{2} - x^{2}}} - \int_{- a}^{a} \frac{xdx}{\sqrt{a^{2} - x^{2}}} = a \cdot 2 \int_{0}^{a} \frac{dx}{\sqrt{a^{2} - x^{2}}} - 0 (∵ \frac{x}{\sqrt{a^{2} - x^{2}}} is an odd function) \\ = 2 a {[\sin^{- 1} \frac{x}{a}]}_{0}^{a} \\ \Rightarrow 2 a [\sin^{- 1} (1) - \sin^{- 1} (0)] = πa \end{array}$