对xcos^2xdx求定积分,上限2派=360'下限0
∫(0到π) (e^xcos�0�5x) dx=∫(0到π) [e^x*1/2*(1+cos2x)] dx=1/2*∫(0到π) e^x dx+1/2*∫(0到π) e^x*cos2x dx=1/2*(e^π-1)+1/2*∫(0到π) e^x*cos2x dx 独立求∫(0到π) e^x*cos2x dx=∫(0到π) e^x d(1/2*sin2x)=e^x*1/2*sin2x(0到π)-1/2*∫(0到π) e^x*sin2x dx,分部积分=-1/2*∫(0到π) e^x d(-1/2*cos2x)=-1/2*-1/2*cos2x*e^x(0到π)-(-1/2)(-1/2)∫(0到π) e^x*cos2x dx,分部积分=1/4*(e^π-1)-1/4*∫(0到π) e^x*cos2x dx(1+1/4)∫(0到π) e^x*cos2x dx=1/4*(e^π-1)∫(0到π) e^x*cos2x dx=1/5*(e^π-1) ∴原式=1/2*(e^π-1)+1/2*[1/5*(e^π-1)]=3/5*(e^π-1)
∫(0→2π) xcos²x dx
= ∫(0→2π) x • (1 + cos2x)/2 dx
= (1/2)∫(0→2π) x dx + (1/2)∫(0→2π) xcos2x dx
= (1/4)[ x² ]:(0→2π) + (1/4)∫(0→2π) x d(sin2x)
= (1/4)(4π²) + (1/4)xsin2x:(0→2π) - (1/4)∫(0→2π) sin2x dx
= π² + 0 + (1/8)cos2x:(0→2π)
= π² + (1/8)(1 - 1)
= π²
如何求∫xcos^2xdx?
方法一:换元积分法 令 u = xcosx, du = (cosx - xsinx) dx,那么:∫xcos^2xdx = ∫u^2\/(u^2 + sin^2x) du = ∫(u^2 + sin^2x - sin^2x)\/(u^2 + sin^2x) du = ∫(1 - sin^2x\/(u^2 + sin^2x)) du = ∫1 du - ∫sin^2x\/(u^2 + sin^2x) du = u ...
对xcos^2xdx求定积分,上限2派=360'下限0
∫(0到2π) x*cos�0�5x dx=∫(0到2π) x*1\/2*(1+cos2x) dx=1\/2*∫(0到2π) x dx+1\/2*∫(0到2π) x*cos2x dx=1\/2*1\/2*x�0�5(0到2π)+1\/2*∫(0到2π) x*cos2x dx 对于∫(0到2π) x*cos2x dx=∫(0到2π) x*d(...
不定积分 :∫ xcos^2xdx 求详细过程和答案 拜托大神.
∫ xcos^2xdx =∫ x(1+cos2x\/2)dx =1\/2∫ xdx+1\/2∫xcos2xdx =x²\/4+1\/4∫xdsin2x =x²\/4+1\/4*xsin2x-1\/4∫sin2xdx =x²\/4+1\/4*xsin2x-1\/8∫sin2xd2x =x²\/4+1\/4*xsin2x+1\/8*cos2x+C ...
求不定积分∫x(cosx)^2dx
∫x(cosx)^2dx=∫xcos^2xdx =∫x(1+cos2x\/2)dx =1\/4x^2+1\/2∫xcos2xdx =1\/4x^2+1\/4∫xd(sin2x)=1\/4x^2+1\/4xsin2x-1\/4∫sin2xdx =1\/4x^2+1\/4xsin2x+1\/8cos2x+C 说明:C是常数 不可积函数 虽然很多函数都可通过如上的各种手段计算其不定积分,但这并不意味着所有...
cos^2x求不定积分
回答:∫cos^2xdx =∫(1+cos2x)dx\/2 =∫(1+cos2x)d2x\/4 =(1\/4)∫[d2x+cos2xd2x] =(1\/4){2x+sin2x+C1} =x\/2+(sin2x)\/4+C
∫(0,2丌)xcos^2xdx
∫(0,2丌)xcos^2xdx 我来答 1个回答 #热议# 为什么现在情景喜剧越来越少了? 百度网友31ba68d 2014-01-08 知道答主 回答量:20 采纳率:0% 帮助的人:12万 我也去答题访问个人页 关注 展开全部 本回答由提问者推荐 已赞过 已踩过< 你对这个回答的评价是? 评论 收起 为你推荐: ...
求x*cos^2x的不定积分
∫x cos2x dx =1\/2∫x (1+cos2x)dx =1\/2∫xdx + 1\/4∫xdsin2x =x2\/4 + 1\/4 x sin2x - 1\/4∫sin2xdx =x2\/4 + 1\/4 x sin2x + 1\/8 cos2x + C 这个不是分步积分法!!是一般的化解法,先三角降次,再用凑微分法。楼上不要误导哦 ...
sin( x+π\/4)可积吗?
方法如下,请作参考:
专升本数学问题
(1)∫xcos2xdx =∫x(1-2sinx^2)dx = ∫xdx- ∫2xsinx^2dx =1\/2(x^2)-∫sinx^2dx^2 (x^2看做一个整体t)=1\/2(x^2)+cosx^2+C (2) ∫(1,0)x√(1-x^2)dx (将x移入dx中变成1\/2dx^2)=1\/2∫(1,0)√(1-x^2)dx^2 =-1\/2∫(1,0)√(1-x...
∫cos²2xdx,高等数学,不定积分
倍角加分步 cos^2x=(cos2x+1)\/2 原因为化为 ∫1\/2*x^2dx+1\/4∫x^2dsin2x =1\/6x^3+1\/4sin2x*x^2-1\/2∫xsin2xdx =1\/6x^3+1\/4sin2x*x^2+1\/4xcos2x-1\/4∫cos2xdx =1\/6x^3+1\/4sin2x*x^2+1\/4xcos2x+1\/8sin2x 思路是这样,错没错不晓得 ...
费盲歧星: ∫(0到2π) x*cos²x dx =∫(0到2π) x*1/2*(1+cos2x) dx =1/2*∫(0到2π) x dx+1/2*∫(0到2π) x*cos2x dx =1/2*1/2*x²(0到2π)+1/2*∫(0到2π) x*cos2x dx 对于∫(0到2π) x*cos2x dx =∫(0到2π) x*d(1/2*sin2x) =1/2*x*sin2x(0到2π)-1/2*∫(0到2π) sin2x dx =0+1/4*cos2x(0到2π) =0 ∴原式=1/2*1/2*x²(0到2π)+0 =π²
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费盲歧星: ∫ x/cos²x dx=∫ xsec²x dx=∫ x d(tanx)=xtanx - ∫ tanx dx=xtanx - ∫ sinx/cosx dx=xtanx + ∫ 1/cosx d(cosx)=xtanx + ln|cosx| + C 【数学之美】团队为您解答,若有不懂请追问,如果解决问题请点下面的“选为满意答案”.
化隆回族自治县18216032674: 求定积分∫x^2cosxdx,上限是2π,下限是0 - ?
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费盲歧星: 这题方法有很多,你可以把 cos^2x换成1-sin^2x 4sin^2xcos^2x =4(sin^2x-sin^4x) sin^2x 和 sin^4x 积分是有公式的.但是一般人估计也记不得,所以方法二:为了方便,上下限不写,最后带 原式=4∫(1-cos2x)/2 * (cos2x+1)/2 dx =∫(1-cos2x)* (cos2...
化隆回族自治县18216032674: 对e^xcos^2xdx求定积分,上限派=180'下限0 - ?
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费盲歧星: ∫根号1+cos2xdx,积分上限是π,积分下限是0的值是2√ ∵cos2x=2cos²x-1 ∴∫√(1+cos2x)dx=∫√2|cosx|dx ∴(0,π)∫√(1+cos2x)dx =(0,π/2)∫√2cosxdx+(π/2,π)∫-√2cosxdx =2√2 所以∫根号1+cos2xdx,积分上限是π,积分下限是0的值是2√2. ...
化隆回族自治县18216032674: 求定积分∫上限π/2下限 - π/2 (1+x)cosx/1+cos^2xdx - ?
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化隆回族自治县18216032674: 求定积分x²cos2xdx上限为π下限为0 - ?
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