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∫1/sin(3x)dx=? thx
步骤+正确答案,谢谢
人气:472 ℃ 时间:2020-02-05 07:06:52
解答
∫1/sin(3x)dx
=1/3∫1/sin(3x)d3x
=1/3∫csc(3x)d3x
现在先来计算:∫cscxdx
∫cscxdx
=∫dx/sinx
=∫sinxdx/sin²x
=-∫dcosx/sin²x
=∫dcosx/(cos²x-1)
=(1/2)[∫dcosx/(cosx-1)-∫dcosx/(cosx+1)]
=(1/2)(ln|cosx-1|-ln|cosx+1|)
=(1/2)ln|(cosx-1)/(cosx+1)|
(对数里分子分母都乘以cosx-1)
=(1/2)ln|(cosx-1)²/sin²x|
=ln|(cosx-1)/sinx|
=ln|cotx-cscx|
所以
∫1/sin(3x)dx
=1/3∫1/sin(3x)d3x
=1/3∫csc(3x)d3x
=1/3ln|cot3x-csc3x|
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