举报
dxk1979
这源于一个正弦函数n次幂的定级分推导: I[n] = [0,π/2]∫(sinx)^n *dx = [0,π/2]∫(cosx)^n *dx 作置换x =π/2 – t,则dx = -dt [0,π/2]∫(sinx)^n *dx = [π/2,0]∫[sin(π/2-t)]^n *(-dt) = [0,π/2]∫(cost)^n *dt 利用分部积分: I[n] = [0,π/2]∫(sinx)^n *dx = [0,π/2]∫(sinx)^(n-1) *(-d cosx) = -cosx *(sinx)^(n-1)|[0,π/2] + (n-1)* [0,π/2]∫(sinx)^(n-2) *cos²x *dx = 0 + (n-1)* [0,π/2]∫(sinx)^(n-2) *(1-sin²x)dx = (n-1)* [0,π/2]∫(sinx)^(n-2) *dx - (n-1)* [0,π/2]∫(sinx)^n *dx = (n-1)I[n-2] –(n-1)*I[n] ==> I[n]= I[n-2] *(n-1)/n; 再由I[0] = [0,π/2]∫dx =π/2,I[1] = [0,π/2]∫sinx *dx = 1;据此递推公式可得: 当n为偶数时: I[n] = (n-1)/n*(n-3)/(n-2)*……*1/2*π/2; 当n为奇数时: I[n] = (n-1)/n*(n-3)/(n-2)*……*2/3*1; 在本题中的中间步骤就来自递推公式的偶数情况: 2c^4 *[0,π/2]∫ sin²a(1-sin²a) *da = 2c^4 *{ [0,π/2]∫ sin²a *cos²a *da - [0,π/2]∫ (sina)^4 *da } = 2c^4 * (1/2*π/2 - 3/4*1/2*π/2) = π/8 * c^4