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B
2π3
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C
5π6
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D
π3
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Solution
The correct option is Cπ3 I=∫552√(25−x2)3x4dx Let x=5sinθ ∴dx=5cosθdθ ∴I=∫π2π653cos3θ.5cosθ54sin4θdθ =∫π2π6cot2θ(cosec2θ−1)dθ =∫π2π6cot2θcosec2θdθ−∫π2π6cot2θdθ =∫π2π6cot2θcosec2θdθ−∫π2π6(cosec2θ−1) =[−cot3θ3+cotθ+θ]π2π6 =−0+0+π2−(−3√33+√3+π6)=π3