Question

方框里面的定积分怎么做？

Answer 1

积分函数 f(x) = sinxcosx/[(sinx)^4+(cosx)^4] = (1/2)sin2x/[1-2(sinxcosx)^2] = (1/2)sin2x/[1-(1/2)(sin2x)^2] = sin2x/[2-(sin2x)^2] = A/(√2-sin2x) + B/(√2+sin2x) = [√2(A+B)+(A-B)sin2x]/[2-(sin2x)^2] A+B = 0, A-B = 1, 解得 A = 1/2, B = -1/2, 积分函数化为 f(x) = (1/2)[1/(√2-sin2x)-1/(√2+sin2x)] I = (1/2) ∫ [1/(√2-sin2x)-1/(√2+sin2x)]dx = (1/4) ∫ [1/(√2-sin2x)-1/(√2+sin2x)]d(2x) 令 tanx = u， 则 sin2x = 2/(1+u^2), d(2x) = 2du/(1+u^2) I = (√2/4)∫du/[u^2-(√2-1)] - (√2/4)∫du/[u^2+(√2+1)] = (√2/4)∫du/[u^2-(√2-1)] - (√2/4)∫du/[u^2+(√2+1)] = {√2/[8√(√2-1)]} ∫{[1/[u-√(√2-1)]-[1/[u+√(√2-1)]}du - (√2/4)∫du/[u^2+(√2+1)] = {√2/[8√(√2-1)]}ln|[u-√(√2-1)]/[u+√(√2-1)]| - {√2/[4√(√2+1)]}arctan[u/√(√2+1)] + C = {√2/[8√(√2-1)]}ln|[tanx-√(√2-1)]/[tanx+√(√2-1)]| - {√2/[4√(√2+1)]}arctan[tanx/√(√2+1)] + C

Answer 2

(1/3)[x^3f(x)]<0, 1> = (1/3)[f(1) - 0] = (1/3)(0 - 0) = 0