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【答案】应助回帖
感谢参与，应助指数 +1
Integral{sqrt[1-Sin(2*x)]*dx,0,π/2}
=1/2*Integral{sqrt[1-Sinu]*du,0,π}
=1/2*Integral{sqrt[1-Cos(π/2-u)]*du,0,π}
=1/2*Integral{abs[Sin(π/4-u/2)]*du,0,π}
=1/2*Integral{Sin(π/4-u/2)*du,0,π/2} +
1/2*Integral{Sin(-π/4+u/2)*du,π/2,π}
=[1-1/sqrt(2)]+[1-1/sqrt(2)]
=2-sqrt(2) |
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回复此楼
