已知f(n)=∫(tanx)∧n dx,上限π/4,下限0，求(1) f(n)+f(n+2)；(2
答案:1 悬赏:30 手机版
解决时间 2021-01-14 17:17
- 提问者网友：富士山上尢
- 2021-01-14 08:20
已知f(n)=∫(tanx)∧n dx,上限π/4,下限0，求(1) f(n)+f(n+2)；(2
最佳答案
- 五星知识达人网友：酒者煙囻
- 2021-01-14 09:54
(1)
f(n)= ∫(0->π/4) (tanx)^n dx
f(n)+f(n+2) =∫(0->π/4) (tanx)^n (secx)^2 dx
=∫(0->π/4) (tanx)^n dtanx
=(1/(n+1))[ (tanx)^(n+1)] |(0->π/4)
= 1/(n+1)
(2)
f(1) =∫(0->π/4) (tanx) dx
= -[ln|cosx|] |(0->π/4)
= (1/2)ln2
f(5)+f(3) = 1/6 (1)
f(3)+f(1) = 1/4 (2)
(1) -(2)
f(5)-f(1) = -1/12
f(5) = (1/2)ln2 -1/12
f(n)= ∫(0->π/4) (tanx)^n dx
f(n)+f(n+2) =∫(0->π/4) (tanx)^n (secx)^2 dx
=∫(0->π/4) (tanx)^n dtanx
=(1/(n+1))[ (tanx)^(n+1)] |(0->π/4)
= 1/(n+1)
(2)
f(1) =∫(0->π/4) (tanx) dx
= -[ln|cosx|] |(0->π/4)
= (1/2)ln2
f(5)+f(3) = 1/6 (1)
f(3)+f(1) = 1/4 (2)
(1) -(2)
f(5)-f(1) = -1/12
f(5) = (1/2)ln2 -1/12
我要举报
如以上回答内容为低俗、色情、不良、暴力、侵权、涉及违法等信息，可以点下面链接进行举报！
点此我要举报以上问答信息
大家都在看
推荐资讯