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0aa2x25/2dx=

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a
2πa632
b
5πa632
c
5πa630
d
3πa635

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detailed solution

Correct option is B

Put x=asin⁡θ⇒dx=acos⁡θdθ G.I =∫0π/2 a2−a2sin2⁡θ5/2(acos⁡θ)dθ=a6∫0π/2 cos5⁡θcos⁡θdθ=a6∫0π/2 cos6⁡θdθ=a6⋅56⋅34⋅12⋅π2=5πa632


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