nynw.net
当前位置:首页 >> ∫√xsin√xDx >>

∫√xsin√xDx

换元法+分部积分

换元

令√x=u,则x=u原式为 ∫usinudu =2 ∫u sinudu =-2 ∫ udcosu =-2(ucosu- ∫ cosud u ) =-2ucosu+4∫ ucosud u =-2ucosu+4∫ udsinu =-2ucosu+4usinu-4∫ sinudu =-2ucosu+4usinu+4cosu 故代换得-2xcos√x+4√xsin√x+4cos√x

显然sin²x= -0.5cos2x+0.5 所以得到 原积分 =∫ x *(-0.5cos2x+0.5) dx =0.25x² - 0.25∫ x d(sin2x) =0.25x² - 0.25x *sin2x + ∫0.25sin2x dx =0.25x² - 0.25x *sin2x -0.125cos2x +C,C为常数

你好,答案如下: 需用分部积分法 ∫ x sin(100x) dx = -1/100*∫ x d(cos(100x)) = -1/100*xcos(100x)+1/100*∫ cos(100x) dx = -1/100*xcos(100x)+1/100^2*sin(100x) + C 很高兴能回答您的提问,您不用添加任何财富,只要及时采纳就是对我们最好...

如下 ∫xsin²xdx =1/2∫x(1-cos2x)dx =x^2/4-1/2∫xcos2xdx =x^2/4-1/4∫xdsin2x =x^2/4-1/4xsin2x+1/4∫sin2xdx =x^2/4-1/4xsin2x-1/8cos2x+C

分部积分:原式=-1/k*∫xd(coskx) =-1/kxcoskx+1/k∫coskxdx =-1/kxcoskx+1/k^2sinkx+C

原式=∫√xsin√x×2√xd√x=∫2xsin√xd√x, 令√x=t,则原式=∫2t^2sintdt =﹣∫2t^2dcost =﹣2t^2cost+2∫costdt^2 =﹣2t^2cost+4∫tcostdt =﹣2t^2cost+4∫tdsint =﹣2t^2cost+4tsint-4∫sintdt =﹣2t^2cost+4tsint+4cost+C =…………

sin(x+a)-xcos(x+a)+C 解: ∫xsin(x+a)dx =-∫xdcos(x+a) =-[xcos(x+a)-cos(x+a)d x] =-[xcos(x+a)-sin(x+a)]+C =sin(x+a)-xcos(x+a)+C

你把上面划线的那个等式看成一个整体,移过去,就可以得出下面的等式

网站首页 | 网站地图
All rights reserved Powered by www.nynw.net
copyright ©right 2010-2021。
内容来自网络,如有侵犯请联系客服。zhit325@qq.com