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求定积分∫(0到π/2)[cos(x/2)-sin(x/2)]^2dx
运用到什么三角变换?
人气:104 ℃ 时间:2020-04-29 18:30:06
解答
[cos(x/2)-sin(x/2)]²=[cos²(X/2)+sin²(x/2)]+2sin(x/2)cos(x/2)=1+sinx∫(π/2,0)[cos(x/2)-sin(x/2)]^2dx=∫(π/2,0) 1+sinx dx=x|(π/2,0)-cosx|(π/2,0)=(π/2)+1
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