作者LuisSantos (希望我备上政大应数所)
看板trans_math
标题Re: [积分] 三角函数的积分
时间Sun May 18 02:04:28 2008
※ 引述《euler3002 (无)》之铭言:
: S dx/1+tanx^(根号二), 范围是0到pi/2
: 完全不知道如何下手,谢谢
π/2 1
∫ ------------------ dx
0 1 + (tanx)^(√2)
π/2 1
= ∫ --------------------------- dx
0 1 + ((sinx)/(cosx))^(√2)
π/2 (cosx)^(√2)
= ∫ ----------------------------- dx
0 (cosx)^(√2) + (sinx)^(√2)
π/2 1 π/2 (cosx)^(√2)
令 I = ∫ ------------------ dx = ∫ ----------------------------- dx
0 1 + (tanx)^(√2) 0 (cosx)^(√2) + (sinx)^(√2)
π/2 (cosx)^(√2)
则 I = ∫ ----------------------------- dx
0 (cosx)^(√2) + (sinx)^(√2)
0 (cos(π/2 - y))^(√2)
= ∫ (-----------------------------------------------)(-1) dy
π/2 (cos(π/2 - y))^(√2) + (sin(π/2 - y))^(√2)
(令 y = π/2 - x , 则 x = π/2 - y , dx = - dy)
(x = 0 => y = π/2 , x = π/2 => y = 0)
π/2 (siny)^(√2)
= ∫ ----------------------------- dy
0 (siny)^(√2) + (cosy)^(√2)
π/2 (sinx)^(√2)
= ∫ ----------------------------- dx
0 (sinx)^(√2) + (cosx)^(√2)
π/2 (sinx)^(√2)
则 2I = ∫ ----------------------------- dx
0 (sinx)^(√2) + (cosx)^(√2)
π/2 (cosx)^(√2)
+ ∫ ----------------------------- dx
0 (sinx)^(√2) + (cosx)^(√2)
π/2 (sinx)^(√2) + (cosx)^(√2)
= ∫ ----------------------------- dx
0 (sinx)^(√2) + (cosx)^(√2)
π/2 |π/2 π
= ∫ 1 dx = x | = ----
0 |0 2
π π/2 1 π
I = --- => ∫ ------------------ dx = ----
4 0 1 + (tanx)^(√2) 4
--
※ 发信站: 批踢踢实业坊(ptt.cc)
◆ From: 61.66.173.21
1F:→ euler3002:谢谢回答 59.112.0.9 05/18 23:52