(2) 令 x = sint，得
I = ∫<0, 1>(1-x^2)^(5/2)dx = ∫<0, π/2>(cost)^6dt
= (1/8) ∫<0, π/2>(1+cos2t)^3dt
= (1/8) ∫<0, π/2>[1+3cos2t+3(cos2t)^2+(cos2t)^3]dt
= (1/8) {∫<0, π/2>[5/2+3cos2t+(3/2)cos4t]dt
+ (1/2)∫<0, π/2>[1-(sin2t)^2]dsin2t}
= (1/8)[5t/2+(3/2)sin2t+(3/8)sin4t+(1/2)sin2t-(1/6)(sin2t)^3]<0, π/2>
= 5π/32
(3) 令√(1+x) = u，得 x = u^2-1, dx = 2udu
I = ∫<1, 2>2(u^2-1)^2du = 2∫<1, 2>(u^4-2u^2+1)du
= 2[u^5/5 - 2u^3/3 + u]<1, 2> = 76/15
