∫(上限3下限0)sint^2dt怎麼算
突然忘記怎麼算了= =
2005-07-28 19:54:13 · 5 個解答 · 發問者 Anonymous in 科學 ➔ 數學
平方
2005-07-28 20:14:06 · update #1
^ 符號代表次方
x^3 x 的 3次方
依此類推
2005-07-31 15:01:55 · answer #1 · answered by Big_John-tw 7 · 0⤊ 0⤋
"^"----->代表平方的意思 在有關數學的程式裡面 常常用到
2005-07-29 15:38:06 · answer #2 · answered by 皇羯 2 · 0⤊ 0⤋
因為 cos2t= cost^2-sint^2=1-2sint^2= 2cost^2-1
現在把cos2t=1-2sint^2移項
得sint^2=(1- cos2t)/2
∫(上限3下限0)sint^2 dt
=∫(上限3下限0)(1- cos2t)/2 dt
= t/2- sin2t/4(上限3下限0)
=〔3/2 - (sin6)/4〕- 〔0/2 - (sin0)/4〕
= 3/2 - (sin6)/4
2005-07-29 05:40:10 · answer #3 · answered by 相遇即是有緣 4 · 0⤊ 0⤋
先算不定積分
∫(sint)^2 dt =∫(1/2)-(1/2)*cos2t dt = (1/2)t-(1/4)*sin2t + C
代入上下限可得 (3/2)-(1/4)*sin6
2005-07-28 23:44:31 · answer #4 · answered by Anonymous · 0⤊ 0⤋
"^"是啥意思?
2005-07-28 19:56:28 · answer #5 · answered by Anonymous · 0⤊ 0⤋