If \( I_n = \int_0^a (a^2 - x^2)^n \, dx \) where \( n \) is a positive integer, then the relation between \( I_n \) and \( I_{n-1} \) is:
Integrate by parts, taking \((a^2 - x^2)^n\) as the first function and 1 as the second:
\(I_n = \left[x(a^2 - x^2)^n\right]_0^a + 2n\int_0^a x^2(a^2 - x^2)^{n-1}\,dx\)
The boundary term is 0. Write \(x^2 = a^2 - (a^2 - x^2)\):
\(I_n = 2n\left[a^2 I_{n-1} - I_n\right] \Rightarrow (2n + 1)I_n = 2na^2 I_{n-1}\)
\(I_n = \frac{2na^2}{2n+1}I_{n-1}\)
Answer: (A)
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