定积分xcosx∧2 xsinx的定积分

=0.5∫sec(x^2)d(x^2) =0.5ln|tan(x^2/2+π/4)|+C

你好!用凑微分法∫xcos(x^2)dx=∫cos(x^2)d(x^2)=sin(x^2)+c.经济数学团队帮你解答,请及时采纳.谢谢!

解: ∫xcos²x dx=∫x(1+cos2x)/2 dx=∫x/2 dx+½∫xcos2xdx=x²/4+1/4·x·sin2x-1/4∫sin2x dx=x²/4+x/4·sin2x+1/8·cos2x+C=1/8·(2x²+2x sin2x+cos2x)+C

首先 Cos(2x) = Cosx ^2 - Sinx^2 = Cosx^2 -1 + Cosx^2=> Cosx^2 = 1/2 ( 1 + Cos (2x)) 所以1/2 积分号:(0 - 2π) x + xcos 2x 这样你会了吧?x直接积即可, 后面部分用分部积分 V＂ = Cos (2x) u = X

x(cosx)^2=1/2*x*(1+cos2x)=x/2+xcos2x/2 然后分别积,分了吧,后面一项用分步积分

(sinx^2 )/2 + C

∫x*(secx)^2 * dx=∫x * d(tanx)=x*tanx - ∫tanx *dx=x*tanx + ∫d(cosx)/cosx=x*tanx + ln|cosx| + C

此题用分部积分法 ∫x/(cosx)^2dx=∫x (secx)^2dx=∫xd(tanx)=xtanx-∫tanxdx=xtanx+ln/cosx/ 最后把积分上下限一代就得到答案了 注:∫tanxdx=∫sinx/cosx dx =-∫d(cosx)/cosx =-ln/cosx/+c

∫xcosx^2dx=(1/2)∫cosx^2dx^2=(1/2)sinx^2+C

∫xcos^2 x dx=∫x(cos2x+1)/2 dx=1/2*∫xcos2xdx+1/2*∫xdx=1/4∫xcos2xd2x+1/4∫dx^2=1/4∫xdsin2x +x^2/4=1/4 *xsin2x-1/4∫sin2xdx +x^2/4=xsin2x/4+x^2/4-1/8∫sin2xd2x=xsin2x/4+x^2/4+1/8∫dcos2x=xsin2x/4+x^2/4+cos2x/8 + C