高数第三题!

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利用对称性：
0≤p≤2cosθ
π/4≤θ≤π/2
原式=2∫∫1/p ·pdpdθ
=2∫∫dpdθ
=2∫(π/4,π/2)dθ∫(0,2cosθ)dp
=2∫(π/4,π/2)2cosθdθ
=4sinθ|(π/4,π/2)
=4sinπ/2-4sinπ/4
=4-4×√2/2
=4-2√2