已知f(cosx)=(sinx)∧2,則∫f(x-1)dx=?

題目：

已知f(cosx)=(sinx)∧2,則∫f(x-1)dx=?

解答：

f(cosx) = sin²x = 1 - cos²x ===> f(x) = 1 - x²
令x -1 = t ====> x = t +1 dx = dt
原式 = ∫f(x-1)dx = ∫f(t)dt = ∫(1 - t²) dt = t - t³/3 + C = (x -1) - (x -1)³/3 + C