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sin根号xDx

你好 令√x=t,x=t ,dx=2tdt 原式=∫sint*2tdt =∫(1-cos2t)/2*2tdt =∫tdt-∫tcos2tdt =1/2t -1/2∫td(sin2t) =1/2t -1/2(tsin2t-∫sin2tdt) =1/2t -1/2tsin2t-1/4cos2t+c =1/2x-1/2√xsin2√x-1/4cos2√x+c 【数学辅导团】为您解答,不理解请追问,理解请及时选为满意回答!(*^__^*)谢谢!

令√x=t∫sin√xdx=2∫tsintdt=-2∫tdcost=-2tcost+2∫costdt=-2tcost+2sint+C=-2√xcos√x+2sin√x+C

设√x=t,x=t^2,dx=2tdt,原式=∫sint*2tdt=2∫t*sintdt=2∫td(-cost)=-2tcost+2∫costdt=-2tcost+2sint+C=-2√xcos√x+2sin√x+C.

x = t^2dx = 2tdt原式=2tsint*dt/t = 2sintdt积分得-2cost+C即结果为-2cos根号x+C

计算过程如下:设√x=t,则x=t^2,dx=2tdt.可以得到:原式=∫sint*2tdt=2∫t*sintdt=2∫td(-cost)=-2tcost+2∫costdt=-2tcost+2sint+C=-2√xcos√x+2sin√x+C(以上C为常数) 扩展资料:不定积分求法:1、积分公式法.直接利用积分公式求出不定

令√x=t 则,x=t^2 那么,dx=d(t^2)=2tdt 所以,原式=∫sint*2tdt=2∫sint*tdt =2∫t*d(-cost) =-2∫t*d(cost) =-2*[t*cost-∫costdt] =-2tcost+2∫costdt =-2tcost+2sint+C 将t=√x代入上式就有:∫sin√x=-2√x*cos(

''代表平方.y=1/根号sinx; y''=1/sinx; sinx=1/y'' x=arcsin(1/y''); 故原函数y=arcsin(1/y'')

请写明白

令根号x=t, 即x=t^2, 积分sintdt^2=积分sint*2tdt后分部积分

∫√xsin√xdx=2∫xsin√xd√x=-2∫xdcos√x=-2xcos√x+2∫cos√xdx=-2xcos√x+4∫√xcos√xd√x=-2xcos√x+4∫√xdsin√x=-2xcos√x+4√xsin√x-4∫sin√xd√x=-2xcos√x+4√xsin√x+4cos√x+C 不断地使用分部积分法即可得到答案.

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