∫[1/(√4-9x^)]2dx

解：由于被积函数的定义域为[-2/3，2/3]，所以设x=[2sin（t）]/3，t属于[0，pi/2)并(pi/2,pi];代入原式得：∫[1/(√4-9x^2）]dx
=∫[1/（2*cost）]dx
=（1/2）*∫[cost/(cost)^2]dx
=(1/2)*∫{cost/[(1-(sint)^2]}dt
=(1/4)*∫{[cost/(1+sint)]+[cost/(1-sint)]}dx
=(1/4)*{[∫d(sint)/(1+sint)]+[∫d(sint)/(1-sint)]}
=(1/4)*{[∫d(sint)/(1+sint)]-[∫d(sint)/(sint-1)]}
=(1/4)*ln[(1+sint)/(1-sint)]+C
=(1/4)*ln{(1+sint)^2/[(1-(sint)^2]}+C
=(1/4)*2*ln{[(1+sint)/cost]的绝对值}+C
=(1/2)*ln[（sect+tant)的绝对值]+C
=(1/2)*ln{(2+3x)/[根号下（4-9x^2)]}+C
注意：根据我们的换元可得sect+tant>0,所以2+3x不必加绝对值符号（实际上也不应该加).