为什么x可以直接变成pai/2
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发布时间:2024-10-22 18:53
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热心网友
时间:2024-10-25 00:29
π/2
解析:
//奇函数在对称区间上的定积分为0//
//偶函数在对称区间上的积极分=2S₁//
//周期性函数例如三角函数,在特殊区间上的定积分,呈现特别性质//
~~~~~~~~~~~~~~
积分区间:[0,π]
令t=x-π/2,则:t∈[-π/2,π/2]
~~~~~~~~~~~~~~
S
=∫(π/2+t)√[sin²(π/2+t)-sin⁴(π/2+t)]d(π/2+t),积分区间[-π/2,π/2]
=∫(π/2+t)√(cos²t-cos⁴t)dt
=∫(π/2)√(cos²t-cos⁴t)dt+∫t√(cos²t-cos⁴t)dt
=∫(π/2)√(cos²t-cos⁴t)dt+0
=(π/2)∫√(cos²t-cos⁴t)dt
=(π/2)∫√(sin²t-sin⁴t)dt
=(π/2)*2∫sintcostdtdt,积分区间:[0,π/2]
=π∫sin2td(2t)
=π*(1/4)(-cos2t)+C
=(π/4)[(-cosπ)-(-cos0)]
=(π/4)*2
=π/2
~~~~~~~~~~~~~
cos²t-cos⁴t
=cos²t(1-cos²t)
=(1-sin²t)sin²t
=sin²t-sin⁴t
热心网友
时间:2024-10-25 00:27
是换元t=π-x,利用正弦性质sinx=sin(π-t)
热心网友
时间:2024-10-25 00:32
π/2
解析:
//奇函数在对称区间上的定积分为0//
//偶函数在对称区间上的积极分=2S₁//
//周期性函数例如三角函数,在特殊区间上的定积分,呈现特别性质//
~~~~~~~~~~~~~~
积分区间:[0,π]
令t=x-π/2,则:t∈[-π/2,π/2]
~~~~~~~~~~~~~~
S
=∫(π/2+t)√[sin²(π/2+t)-sin⁴(π/2+t)]d(π/2+t),积分区间[-π/2,π/2]
=∫(π/2+t)√(cos²t-cos⁴t)dt
=∫(π/2)√(cos²t-cos⁴t)dt+∫t√(cos²t-cos⁴t)dt
=∫(π/2)√(cos²t-cos⁴t)dt+0
=(π/2)∫√(cos²t-cos⁴t)dt
=(π/2)∫√(sin²t-sin⁴t)dt
=(π/2)*2∫sintcostdtdt,积分区间:[0,π/2]
=π∫sin2td(2t)
=π*(1/4)(-cos2t)+C
=(π/4)[(-cosπ)-(-cos0)]
=(π/4)*2
=π/2
~~~~~~~~~~~~~
cos²t-cos⁴t
=cos²t(1-cos²t)
=(1-sin²t)sin²t
=sin²t-sin⁴t
热心网友
时间:2024-10-25 00:28
是换元t=π-x,利用正弦性质sinx=sin(π-t)