解：原式=∫<0.π/4>dθ∫<0,sinθ/cos²θ>r*rdr (做极坐标变换)
=∫<0.π/4>(1/3)(sinθ/cos²θ)³dθ
=(1/3)∫<0.π/4>sin³θdθ/(cosθ)^6
=(-1/3)∫<0.π/4>[(cosθ)^(-6)-(cosθ)^(-4)]d(cosθ)
=(-1/3)[(-1/5)(cosθ)^(-5)+(1/3)(cosθ)^(-3)]│<0.π/4>
=(-1/3)[(-1/5)(1/√2)^(-5)+(1/3)(1/√2)^(-3)+1/5-1/3]
=(-1/3)(-4√2/5+2√2/3-2/15)
=2(√2+1)/45。
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