∫xcos∧2xdx

如题所述

第1个回答 2016-02-17

=∫x(1+cos2x)/2dx=x^2/4+∫xcos2x/4d(2x)=x^2/4+∫x/4d(sin2x)
=x^2/4+(xsin2x)/4-1/4∫(sin2x)dx
=x^2/4+(xsin2x)/4-1/8∫(sin2x)d(2x)
=x^2/4+(xsin2x)/4+1/8cos(2x)+C本回答被网友采纳

第2个回答 2016-02-17

x^2/4 + 1/8 Cos[2 x] + 1/4 x Sin[2 x]追问

求过程谢谢

相似回答

不定积分 :∫ xcos^2xdx 求详细过程和答案拜托大神.答：=∫ 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

∫xcos∧2xdx答：=∫x(1+cos2x)/2dx=x^2/4+∫xcos2x/4d(2x)=x^2/4+∫x/4d(sin2x)=x^2/4+(xsin2x)/4-1/4∫(sin2x)dx =x^2/4+(xsin2x)/4-1/8∫(sin2x)d(2x)=x^2/4+(xsin2x)/4+1/8cos(2x)+C

求∫xcos^2xdx等于多少,上限是2π,下限是0.谢谢!!答：= ∫(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...

高数求助答：(1)∫xcos^2xdx =∫x·(1+cos2x)/2 dx =1/2·(∫xdx + ∫xcos2x dx)=1/4·[x^2 + ∫xd(sin2x)]=1/4·x^2 + 1/4·x·sin2x - 1/4·∫d(sin2x) dx =1/4·x^2 + 1/4·x·sin2x + 1/8·cos2x + C (2)∫ln(1+x^2)dx =x·ln(1+x^2) -∫xd[ln(...

∫xcos∧2xdx要过程哦答：2 2012-11-21 不定积分 :∫ xcos^2xdx 求详细过程和答案拜托大... 33 2013-01-17 ∫cos^2xdx。过程谢谢 2012-01-01 求助:∫x^2cos^2xdx 4 2015-03-27 ∫xcos²xdx 52 2015-12-07 ∫x²cos²x/2dx的不定积分,要过... 12 2014-12-25 ∫x/(cos∧2)(1-4x*x) dx的...

对xcos^2xdx求定积分,上限2派=360'下限0答：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*x�0�5(0到2π)+0=π�0�5 ...

∫xcos^2xdx 的积分答：(x*sin(2*x))/4 - sin(x)^2/4 + x^2/4

大家正在搜

∫xcos2xdx定积分 ∫e^xcosxdx ∫arccosxdx ∫xcos2x ∫cot^2xdx ∫cosxdx ∫x2arctanxdx ∫xtan^2xdx ∫xe^2xdx