✔ 最佳答案
a) let u = a-x, ∴ du = -dxwhen x = a, u = 0when x = 0, u = a∫f(a-x) dx upper limit is a lower limit is 0 = ∫f(u) (-du) upper limit is 0 lower limit is a= ∫f(u) du upper limit is a lower limit is 0= ∫f(x)dx upper limit is a lower limit is 0 b)by substituting u = π/2 - x∫[(cosx)^6]/{[(cosx)^6]+[(sinx)^6]} dxupper limit is (π/2) lower limit is 0=∫[(sinx)^6]/{[(cosx)^6]+[(sinx)^6]} dxupper limit is (π/2) lower limit is 0=∫[(cosx)^6]/{[(cosx)^6]+[(sinx)^6]} dxupper limit is (π/2) lower limit is 0 = 1/2 ∫{[(cosx)^6]+[(sinx)^6]} /{[(cosx)^6]+[(sinx)^6]} dx upper limit is (π/2) lower limit is 0 = 1/2 ∫dxupper limit is (π/2) lower limit is 0 = π/4