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aaaxa+xdx is equal to

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a
π
b
a
c
aπ2
d

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detailed solution

Correct option is D

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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