Question

1/x√(1-x^2) 的不定积分

1/cos^2(x)sin^2(x)的不定积分

两题，求解详细过程

Answer 1

（1）解：令x=sint,dx=costdt
∫dx/[x√(1-x^2)]=∫costdt/[sintcost]=∫1/sint*dt=∫sint/(sint)^2*dt=∫sint/[1-(cost)^2]*dt
=-∫d(cost)/[1-(cost)^2]=-1/2*∫[1/(cost+1)-1/(cost-1)]d(cost)
=-1/2*(ln|cost+1|-ln|cost-1)+C
=-1/2*ln|(cost+1)/(cost-1)|+C
=-ln|(sint/(cost-1)|+C
=-ln|x/[√(1-x^2)-1]|+C
=ln|[√(1-x^2)-1]/x|+C
（2）解法1：
解：原式= ∫{ [(sin x)^2 +(cos x)^2 ] /[(sin x)^2 (cos x)^2 ] }dx
= ∫[ (sec)^2 ]dx +∫[ (csc)^2 ]dx
= tan x -cot x +C
= sin x /cos x -cos x /sin x +C
= [ (sin x)^2 -(cos x)^2 ] / (cos x sin x) +C
= -cos 2x / [ (1/2)sin 2x ] +C
= -2 cot 2x +C, (C为任意常数).
解法2：

原式= ∫dx / [(1/4) (sin 2x)^2]
= 4 ∫[ (csc 2x)^2 ] dx
= 2 ∫[ (csc 2x)^2 ] d(2x)
= -2 cot 2x +C, (C为任意常数).

满意请采纳，谢谢~