∫cos^2xdx。过程 谢谢
原式=∫(1+cos2X)/2 dx =∫1/2dx +1/2∫cos2Xdx =1/2X + 1/4∫cos2Xd2x =1/2X + 1/4sin2x + C
∫ 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
=(1/2)∫(cos2x+1)dx
=(1/2)[ sin2x /2 + x] + C
求解∫sin^3xcos^2xdx
2017-04-05 ∫sin^3cos^2xdx = 3 2017-02-25 求cos^2x\/sin^3x的不定积分。 20 2015-11-26 ∫sin^2xcos^2xdx 9 2015-12-26 求解简单高数不定积分∫sin^2x\/cos^3xdx 9 2015-01-10 求不定积分cos^3Xsin^2Xdx 1 2017-01-27 跪求大神解答高数∫sin^2xcos^2×3xdx 1 2018-01-03...
求助:∫x^2cos^2xdx
解:原式=∫x²(1+cos(2x))\/2dx =x³\/6+1\/2∫x²cos(2x)dx =x³\/6+x²sin(2x)\/4-1\/2∫xsin(2x)dx (应用分部积分法)=x³\/6+x²sin(2x)\/4+xcos(2x)\/4-1\/4∫cos(2x)dx (应用分部积分法)=x³\/6+x²sin(2x)\/4+x...
求不定积分∫sinx^3xcox^2xdx的求解过程
∫sin^3x cos^2xdx= - ∫(1-cos^2x)cox^2xdcosx = ∫(cos^4x-cos^2x)dcosx =(1\/5)cos^5x - (1\/3)cos^3x
∫x^2cos^2xdx
简单分析一下,答案如图所示
不定积分∫sinxcos^2xdx分部积分法
∫sinxcos^2xdx=∫sinxcosxdsinx=1\/2∫cosxdsin^2x =1\/2cosxsin^2x+1\/2∫sin^3x=1\/2cosxsin^2x+1\/2∫sinx(1-cos^2x)dx =1\/2cosxsin^2x-1\/2cosx-1\/2∫sinxcos^2xdx 故:∫sinxcos^2xdx=2\/3[1\/2cosxsin^2x-1\/2cosx]+C ...
关于不定积分和定积分
=cos^5xsinx|(0,π\/2)+∫(0,π\/2) 5sin^2xcos^4xdx =∫(0,π\/2) 5(1-cos^2x)cos^4xdx =5∫(0,π\/2) cos^4xdx-5∫(0,π\/2) cos^6xdx ∫(0,π\/2) cos^6xdx=(5\/6)*∫(0,π\/2) cos^4xdx =(5\/6)*(3\/4)*∫(0,π\/2) cos^2xdx =(5\/6)*(3\/4)*(1...
不定积分∫sinxcos^2xdx分部积分法
∫sinxcos^2xdx=∫sinxcosxdsinx=1\/2∫cosxdsin^2x =1\/2cosxsin^2x+1\/2∫sin^3x=1\/2cosxsin^2x+1\/2∫sinx(1-cos^2x)dx =1\/2cosxsin^2x-1\/2cosx-1\/2∫sinxcos^2xdx 故:∫sinxcos^2xdx=2\/3[1\/2cosxsin^2x-1\/2cosx]+C ...
x\/2cos^2xdx怎么求
解:原式=∫x²(1+cos(2x))\/2dx =x³\/6+1\/2∫x²cos(2x)dx =x³\/6+x²sin(2x)\/4-1\/2∫xsin(2x)dx (应用分部积分法)=x³\/6+x²sin(2x)\/4+xcos(2x)\/4-1\/4∫cos(2x)dx (应用分部积分法)=x³\/6+x²sin(2x)\/4+xcos...
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(...
请问大佬不定积分∫-sin^2x\/cos^2xdx怎么求
不定积分∫-sin^2x\/cos^2xdx怎么求,其求法见上图。这道不定积分∫-sin^2x\/cos^2xdx,求时用三角公式变形后,再用积分公式,就可以将这个不定积分求出来了。具体的求不定积分∫-sin^2x\/cos^2xdx的步骤见上。
江茂愈舒:[答案] ∫cos^2 xdx =∫cos2x+1)/2dx =1/4*Scos2xd2x+1/2*Sdx =1/4*sin2x+x/2+c
赣榆县15232898990: 求∫cos^2√xdx详细解答过程!谢谢 - ?
江茂愈舒:[答案] 设 √x=t 原式=2∫tcos2tdt=∫t(cos2t+1)dt =1/2 ∫ td(sin2t) + ∫ tdt =1/2[tsin2t- ∫ sin2tdt]+t2/2 =1/2[tsin2t+1/2cos2t]+t2/2+C, 将t换回x即可.
赣榆县15232898990: ∫cos2xdx 和∫cos^2xdx怎么解啊! - ?
江茂愈舒:[答案] ∫cos2xdx = ∫(1/2)cos2xd(2x )=1/2 ∫cos2xd2x =1/2(sin 2x + C) ∫cos^2xdx=∫(cos4x +1) /2 dx =∫(cos4x +1) /8 d(4x) =1/8 ∫(cos4x +1) d(4x) =1/8 [∫cos4x d(4x)+∫1d(4x)] =1/8 [(sin 4x +C1)+ 4x+C2] =1/8(sin 4x+ 4x+C)
赣榆县15232898990: ∫cos^2xdx.过程 谢谢 - ?
江茂愈舒: ∫cos^2xdx=(1/2)∫(cos2x+1)dx=(1/2)[ sin2x /2 + x] + C
赣榆县15232898990: 求不定积分∫COS^2根号xdx - ?
江茂愈舒: 令t=根号x 再将cos^2=(1+cos2x)/2带入就会得到(1/2)t^2-1/4|cos2t*tdt如果我没算错,最终得到(1/2)t^2+(t/2)sin2t+(1/4)cos2t+c
赣榆县15232898990: 求不定积分 ∫xln^2xdx - ?
江茂愈舒: ∫xln^2xdx=1/2*x²ln²x-1/2*x²lnx+1/4*x²+C.C为积分常数.解答过程如下:∫xln^2xdx=1/2∫ln²xdx²=1/2*x²ln²x-1/2∫x²dln²x=1/2*x²ln²x-1/2∫x²*2lnx*1/xdx=1/2*x²ln²x-1/2∫lnxdx²=1/2*x²ln²x-(1/2*x²lnx-1/2∫x²dlnx)=1/2*x²ln²x-1/...
赣榆县15232898990: 用凑微分法计算定积分cos^2xdx - ?
江茂愈舒: ∫cos^2xdx =∫(1+cox2x)/2dx =1/2∫dx+1/4∫cos2xd(2x)=1/2*x+1/4*sin2x+C
赣榆县15232898990: 不定积分∫sinxcosx^2xdx - ?
江茂愈舒:[答案] ∫sinxcosx^2xdx =-∫cos^2xdcosx =-1/3*cos^3x+C
赣榆县15232898990: ∫sinxcos^2xdx= 急用啊!!!! - ?
江茂愈舒: ∫sinxcos^2xdx=- ∫cos^2xdcosx= 1/3cos^3x
赣榆县15232898990: 用凑微分法计算定积分cos^2xdx - ?
江茂愈舒:[答案] ∫cos^2xdx =∫(1+cox2x)/2dx =1/2∫dx+1/4∫cos2xd(2x) =1/2*x+1/4*sin2x+C