Question

求解一个二重积分的提，题目

Answer 1

解：原式=∫<0,π/2>dθ∫<1/(cosθ+sinθ),1>[(rcos+rsin)/r²]rdr (应用极坐标变换)
=∫<0,π/2>dθ∫<1/(cosθ+sinθ),1>dr
=∫<0,π/2>[1-1/(cosθ+sinθ)]dθ
=∫<0,π/2>[1-1/(√2sin(θ+π/4))]dθ
=∫<0,π/2>[1-(sin(θ+π/4)/(2√2))(1/(cos(θ+π/4)-1)-1/(cos(θ+π/4)+1))]dθ
=[θ-(1/(2√2))ln│(cos(θ+π/4)-1)/(cos(θ+π/4)+1)│]│<0,π/2>
=π/2-(1/(2√2))[ln│(√2+1)/(√2-1)│-ln│(√2-1)/(√2+1)│]
=π/2-ln(√2+1)/√2