已解决 求定积分1/(sinx+cosx)dx积分区间0到1/2派详细解答
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∫<0, π/2>dx/(sinx+cosx) = (1/√2)∫<0, π/2>d(x+π/4)/sin(x+π/4)
= (1/√2)[ln{csc(x+π/4)-cot(x+π/4)}]<0, π/2>
= (1/√2)[ln(√2+1)-ln(√2-1)] = √2ln(√2+1)
= (1/√2)[ln{csc(x+π/4)-cot(x+π/4)}]<0, π/2>
= (1/√2)[ln(√2+1)-ln(√2-1)] = √2ln(√2+1)
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