∫sin(x+π/3)dx等于多少,上限为π,下限为π/3,求详的细解题步骤,O(∩_∩)O谢谢!
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∫sin(x+π/3)dx等于多少,上限为π,下限为π/3,求详的细解题步骤,O(∩_∩)O谢谢!
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∫(π/3,π) sin(x+π/3)dx
=∫(π/3,π) sin(x+π/3)d(x+π/3)
= -cos(x+π/3)|(2π/3,4π/3)
= -cos5π/3+cosπ
=-3/2
再问: 答案是0啊
再答: ∫(π/3,π) sin(x+π/3)dx =∫(π/3,π) sin(x+π/3)d(x+π/3) = -cos(x+π/3)|(π/3,π)(这里一激动变限了。不好意思。。) = -cos2π/3+cos4π/3 =0
再问: ∫(π/3,π) sin(x+π/3)dx,前面的(π/3,π)是什么得来的啊?上限不是π吗?
再答: 唔 我只是为了写法简单而已 括号内前一个是下限,后一个是上限
=∫(π/3,π) sin(x+π/3)d(x+π/3)
= -cos(x+π/3)|(2π/3,4π/3)
= -cos5π/3+cosπ
=-3/2
再问: 答案是0啊
再答: ∫(π/3,π) sin(x+π/3)dx =∫(π/3,π) sin(x+π/3)d(x+π/3) = -cos(x+π/3)|(π/3,π)(这里一激动变限了。不好意思。。) = -cos2π/3+cos4π/3 =0
再问: ∫(π/3,π) sin(x+π/3)dx,前面的(π/3,π)是什么得来的啊?上限不是π吗?
再答: 唔 我只是为了写法简单而已 括号内前一个是下限,后一个是上限
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