Re: [積分] 幾題積分請教
※ 引述《c1986 (可憐重考生)》之銘言:
: 3
: 1. ∫ x^2√(9-x^2)dx
: 0
Let x=3sin(u) , √(9-x^2)= 3cos(u) , then dx= 3cos(u)du , and so
3 π/2
∫ x^2√(9-x^2)dx = 81 ∫ (sin(u))^2 * (cos(u))^2 du
0 0
= 81*β( 3/2 , 3/2 )
Γ(3/2)Γ(3/2)
= 81 * (1/2) ------- = (81/2)*(π/8) = 81π/ 16
Γ(3)
: x
: 2 ∫---------- dx
: √(x^2-2x)
Let (x-1)=sec(u) , √(x^2-2x) = tan(u) , then dx = sec(u)*tan(u)du , and so
x sec(u) + 1
∫---------- dx = ∫-------sec(u)*tan(u)du
√(x^2-2x) tan(u)
= ∫((sec(u))^2 + sec(u))du = tan(u) + ln|tan(u) + sec(u) | + const.
= √(x^2-2x) + ln| (x-1) + √(x^2-2x) | + const.
: 小弟頭昏昏 腦頓頓 轉不太過來
: 指點我一下吧 感激不盡!
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※ 編輯: Fubini 來自: 61.229.163.217 (04/24 02:30)
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