大学数学分析求不定积分

Question

在线等急

Answer 1

先把正切用正余弦表示出来，然后分子分母用积化和差公式，变成余弦的和或者差

换元之后得到这个式子，这个式子是可以计算的，最后结果是

追答

有更简单的方法

第二个式子右边提出个－1，不定积分就是x

然后下一步

最后一步就可以求不定积分了，分子是分母的导数，凑微分就可以了

追问

谢谢

Answer 2

I = ∫dx/[tan(x+1)tan(x+2)]
= ∫2cos(x+1)cos(x+2)dx/[2sin(x+1)sin(x+2)]
= -∫[cos(2x+3)+cos1]dx/[cos(2x+3)-cos1]
= -∫[cos(2x+3)-cos1+2cos1]dx/[cos(2x+3)-cos1]
= -∫{1+2cos1/[cos(2x+3)-cos1]}dx
= - x - cos1∫d(2x+3)/[cos(2x+3)-cos1]
= - x - cos1∫du/(cosu-cos1), u = 2x+3
令 tan(u/2) = t, 则 du = 2dt/(1+t^2), cosu = (1-t^2)/(1+t^2),
I1 = ∫du/(cosu-cos1) = ∫[2dt/(1+t^2)]/[(1-t^2)/(1+t^2)-cos1]
= ∫2dt/[(1-t^2))-cos1(1+t^2)] = ∫2dt/[1-cos1-(1+cos1)t^2]
= [1/√(1-cos1)]∫[1/[√(1-cos1)+t√(1+cos1)] + [1/[√(1-cos1)-t√(1+cos1)]dt
= (1/a)∫[1/(a+bt) + 1/(a-bt)]dt [a = √(1-cos1), b = √(1+cos1)]
= [1/(ab)] [ln|a+bt|-ln|a-bt|] + C = [1/(ab)]ln|(a+bt)/(a-bt)| + C
则 I = - x - [cos1/(ab)]ln|(a+bt)/(a-bt)| + C
= - x - [cos1/(ab)]ln|[a+btan(u/2)]/[a-btan(u/2)]| + C
= - x - [cos1/(ab)]ln|[a+btan(x+3/2)]/[a-btan(x+3/2)]| + C
其中 a = √(1-cos1), b = √(1+cos1)