∫√(x^2-4)dx∫√(x-x^2)dx请用换元法做,分开的.

Question

∫√(x^2-4)dx ∫ √(x-x^2)dx
请用换元法做,
分开的.

Answer 1

1）∫√(x^2-4)dx
令x=2sect,√(x^2-4)=2tant,dx=2secttantdt
∫√(x^2-4)dx=∫2tant2secttantdt
=4∫tantdsect
=4tantsect-4∫(sect)^2sectdt
=4secttant-4∫[(tant)^2+1]sectdt
=4secttant-4∫[(tant)^2sectdt -4∫sectdt
=4secttant-4∫tantdsect-4ln|sect+tant|
即4∫tantdsect=4secttant-4∫tantdsect-4ln|sect+tant|
4∫tantdsect=2（secttant-ln|sect+tant|）+c
=2(x/2*[√(x^2-4)]/2-ln|x/2+*[√(x^2-4)]/2| + c
∫√(x^2-4)dx =1/2x*[√(x^2-4)]-2ln|x/2+*[√(x^2-4)]/2| + c

2）x-x^2=1/4-(x-1/2)^2
令x=1/2+1/2sint,则√(x-x^2)=1/2cost,dx=1/2costdt
∫ √(x-x^2)dx=∫ 1/2cost*1/2costdt=1/4(t+1/2sintcost)+c
=1/4[arcsin(x-1/2)+(x-1/2)*√(x-x^2)]+c