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If In=∫xn√a2−x2dx and (n+k)⋅In=−xn−1(a2−x2)p+(n−1)a2⋅In−2, then 3k−2p=(where m,n∈N;m,n≥2)

Answer» If In=xna2x2dx and (n+k)In=xn1(a2x2)p+(n1)a2In2, then 3k2p=

(where m,nN;m,n2)




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