如图：
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证明题。。。。
教我如何积分足矣
高中数学log 急！
答案:2 悬赏:50 手机版
解决时间 2021-02-05 14:37
- 提问者网友：玫瑰园
- 2021-02-04 15:10
最佳答案
- 五星知识达人网友：西岸风
- 2021-02-04 15:26
it's not hard.
i will tell you the steps.
first split the denominator into (3c+s)(3c-s)
then integrate with partial fractions.
then everything becomes easy to handle.
get it? the key is PARTIAL FRACTION
the result i get is
1/6{log[3cosx+sinx]-log[3cosx-sinx]}
and you simply plug it in.
good luck with your problem set.
i will tell you the steps.
first split the denominator into (3c+s)(3c-s)
then integrate with partial fractions.
then everything becomes easy to handle.
get it? the key is PARTIAL FRACTION
the result i get is
1/6{log[3cosx+sinx]-log[3cosx-sinx]}
and you simply plug it in.
good luck with your problem set.
全部回答
- 1楼网友：何以畏孤独
- 2021-02-04 16:28
1/（3-根号5） 上下同时乘以3+根号5
=(3+根号5)/4
a=(3+根号5)/4-1=(根号5-1)/4
lg(2a)(2a+1)
=lg[(根号5-1)/2×(根号5+1)/2]
=lg[(5-1)/4]
=lg1
=0
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