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Q.

∫−aa a−xa+xdx is equal to

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a

π

b

a

c

aπ2

d

answer is D.

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Detailed Solution

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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