求定积分问题
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∫(0->π/2) dθ ∫(1/(sinθ+cosθ)->2/(sinθ+cosθ) ) dr
=∫(0->π/2) [ 1/(sinθ+cosθ) ]dθ
=∫(0->π/2) 1/(√2sin(θ+π/4)) dθ
=(1/√2) ∫(0->π/2) csc(θ+π/4) dθ
=(1/√2) [ln|csc(θ+π/4)-cot(θ+π/4)|]| (0->π/2)
=(1/√2) [ ln(√2+1) -ln(√2-1) ]
=(1/√2) [ 2ln(√2+1)]
=√2.ln(√2+1)
=∫(0->π/2) [ 1/(sinθ+cosθ) ]dθ
=∫(0->π/2) 1/(√2sin(θ+π/4)) dθ
=(1/√2) ∫(0->π/2) csc(θ+π/4) dθ
=(1/√2) [ln|csc(θ+π/4)-cot(θ+π/4)|]| (0->π/2)
=(1/√2) [ ln(√2+1) -ln(√2-1) ]
=(1/√2) [ 2ln(√2+1)]
=√2.ln(√2+1)
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