Question

下面两道不定积分怎么求

Answer 1

2) 可以用三角函数的万能代换：u = tan x/2
sin x = 2sinx/2 cosx/2 = 2u/(1+u^2)
du = (1/2) sec^2 (x/2) dx = (1+u^2)/2 dx
原积分 = ∫1/[2u^2-5u+2] du
= ∫1/[(2u-1)(u-2)] du
= (-1/3)∫2/(2u-1) - 1/(u-2)] du (partial fraction)
= -(1/3)[ln|2u-1| - ln|u-1| + c
= -(1/3)[ln|2tanx/2 - 1| - ln|tanx/2 - 1| + c
3) 原积分 = ∫1/√[(3)(4/3^2 - (x+1/3)^2] dx (completing the square)
= (1/√3)arcsin[3(x+1/3)/2] + c

Answer 2