Question

大一高数不定积分

例八，谢谢啦

Answer 1

书上不是有答案么。。。
令u = tan(x/2)、dx = 2/(1 + u²) du、sinx = 2u/(1 + u²)
∫ 1/(3 + sinx) dx
= ∫ 1/[3 + 2u/(1 + u²)] * 2/(1 + u²) du
= ∫ (1 + u²)/[3(1 + u²) + 2u] * 2/(1 + u²) du
= 2∫ 1/(3u² + 2u + 3) du
= 2∫ 1/[3(u + 1/3)² + 8/3] du
= (2/3)∫ 1/[(u + 1/3)² + 8/9] du
= (2/3) * √(9/8) * arctan[(u + 1/3) * √(9/8)] + C
= (2/3) * 3/(2√2) * arctan[(3u + 1)/(2√2)] + C
= (1/√2)arctan[(3tan(x/2) + 1)/(2√2)] + C
= (1/√2)arctan[(3sinx + cosx + 1)/(2√2 + 2√2 cosx)] + C

追问

嘿嘿，就想看看有没有其他解法，书上这个没用过

谢谢你

为什么，sinx=2t/1+t^2

追答

sin(2x)
=2sinxcosx
=[2sinxcosx]/[sin²x＋cos²x]
=[2tanx]/[1＋tan²x] 【分子分母同除以cos²x】
则：sinx=(2u)/(1＋u²)

Answer 2

let

t= tan(x/2)
dt = (1/2)[ sec(x/2)]^2 dx
dx = 2dt/(1+t^2)
sinx = 2sin(x/2).cos(x/2) =2t/(1+t^2)
∫ dx/(3+sinx)
=∫ [2dt/(1+t^2)]/[3 +2t/(1+t^2) ]
=2∫ dt/ (3t^2+2t+3)
=(2/3) ∫ dt/[ t^2+(2/3)t+1]
=(2/3)(3√2/4)arctan[(3t+1)/(2√2)] +C
=(√2/2)arctan[(3t+1)/(2√2)] +C
=(√2/2)arctan{ [3tan(x/2)+1]/(2√2) } +C
consider
t^2+(2/3)t+1 = (t+ 1/3)^2 + 8/9
let
t+1/3 = (2√2/3) tanu
dt =(2√2/3) (secu)^2 du
∫ dt/[ t^2+(2/3)t+1]
=∫ (2√2/3) (secu)^2 du/[ (8/9) (secu)^2 ]
= (3√2/4)u +C
= (3√2/4)arctan[(3t+1)/(2√2)] +C

追问

为什么sinx=2t/1+t^2

追答

sin(x/2) = t/√(1+t^2)
cos(x/2)= 1/√(1+t^2)
sinx = 2sin(x/2) .cos(x/2) = 2t/(1+t^2)

Answer 3

首先，奇函数在对称区间的积分值为0，因此该积分的第二部分为0；第一部分积分，被积函数表示x轴上方的半圆 该积分的值等于该半圆的面积。因此 这个积分=1/2*π*2^2+0=2π