求不定积分∫x-2/（x^2+2*x+3）dx为什么limn->无穷π/n（cos^2π/n+cos^22π/n+...+cos^2(n-1)π/n）相当于(cosx)^2在0到π的定积分?

Question

求不定积分∫ x-2/（x^2+2*x+3） dx
为什么
lim n->无穷 π/n （cos^2 π/n+cos^2 2π/n+...+cos^2 (n-1)π/n）
相当于 (cosx)^2在0到π的定积分?

Answer 1

∫ (x - 2)/(x² + 2x + 3) dx
= ∫ [(1/2)(2x + 2 - 2) - 2]/(x² + 2x + 3) dx
= (1/2)∫ (2x + 2)/(x² + 2x + 3) dx - 3∫ dx/(x² + 2x + 3)
= (1/2)∫ d(x² + 2x + 3)/(x² + 2x + 3) - 3∫ dx/[(x + 1)² + 2]
= (1/2)ln|x² + 2x + 3| - (3/√2)arctan[(x + 1)/√2] + C
对的,定积分的基本定义.
∫(0->π) cos²x dx
= lim(n->∞) π/n * [cos²(π/n) + cos²(2π/n) + ...+ cos²((n - 1)π/n)]
= lim(n->∞) π/n * Σ(k = 1^(n-1)) cos²(kπ/n)
= lim(n->∞) π/n * (n-2)/2
= lim(n->∞) π/2 * (n-2)/n
= lim(n->∞) π/2 * (1-2/n)
= π/2