1/x√(1-x^2)的不定积分1/cos^2(x)sin^2(x)的不定积分

侯锡九回答：

　　令x=sint,dx=costdt∫dx/[x√(1-x^2)]=∫costdt/[sintcost]=∫1/sint*dt=∫sint/(sint)^2*dt=∫sint/[1-(cost)^2]*dt=-∫d(cost)/[1-(cost)^2]=-1/2*∫[1/(cost+1)-1/(cost-1)]d(cost)=-1/2*(ln|cost+1|-ln|cost-1)+...