y'=p,p'=p/x+sinp/xu=p/x,p=ux,p'=u+u'x u+u'x=u+sinudu/sinu=dx/xln(tanu/2)=lnx+lnC1tanu/2=C1xu=2arctan(C1x)y'=p=ux=2xarctan(C1x)y=∫2xarctan(C1x)dx=∫arctan(C1x)dx^2=x^2arctan(C1x)-∫C1x^2/(1+(C1x)^2)=x...du/sinu=dx/x到ln(tanu/2)=lnx lnc1是怎么弄出来的两边积分左边是怎么积出tan的，还是u/2求解释∫du/sinu=∫du/(2sinu/2cosu/2)=∫du/(2tanu/2(cos(u/2))^2=∫(1/tanu/2)dtanu/2=ln|tanu/2|+c