Question

第二换元法求不定积分！！重谢第八题！！

Answer 1

(8)∫√[(1-x)/x] dx
令ψ=√[(1-x)/x], 则ψ²=(1-x)/x => x=1/(ψ²+1) => dx=-2ψ/(ψ²+1)² dψ
=> -2∫ψ²/(ψ²+1)² dψ
令ψ=tanω => dψ=sec²ω dω
sinω=ψ/√(ψ²+1)，cosω=1/√(ψ²+1)
=> -2∫tan²ω/(tan²ω+1)² * (sec²ω dω)
=> -2∫tan²ω/sec⁴ω * (sec²ω dω)
=> -2∫tan²ω/sec²ω dω
=> -2∫sin²ω dω
=> ∫(cos2ω-1) dω
=> sinωcosω - ω + C
=> [ψ/√(ψ²+1)][1/√(ψ²+1)] - arccos[1/√(ψ²+1)] + C
=> √[(1-x)/x]/[(1-x)/x + 1] - arccos{1/√[(1-x)/x + 1]} + C
=> √(x-x²) - arccos√x + C
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