Question

∫1/[(4-9X^2)^1/2]dx

详解...谢谢

Answer 1

令x＝2/3*sinA,则dx=2/3*dsinA=2/3*cosAdA
所以：∫1/[(4-9X^2)^1/2]dx＝∫[1/2cosA*2/3*cosAdA
=∫1/3dA=1/3*A+C=1/3*arcsin1.5x+C

Answer 2

积分:dx/根号(x^2-a^2)(a>0)
令x=asect,0<t<pai/2
原式=积分:asecttantdt/atant
=积分:sectdt
=ln|sect+tant|+C
=ln|x/a+根号(x^2-a^2)/a|+C
=ln|x+根号(x^2-a^2|+C
∫1/(9x^2-4)^(1/2)dx
=积分:1/((3x)^2-2^2)dx
=1/3积分:1/(3x)^2-2^2)d3x
=1/3ln|3x+根号(9x^2-4|+C

Answer 3

根号下提出4，令(3x/2)的平方等于正弦的平方。再算就知道了，看高数不定积分自变量代换。

Answer 4

原式=∫1/[2(1-9X^2/4)^1/2 dx
=1/2∫1/(1-9X^2/4)^1/2 dx
=1/3∫1/(1-9X^2/4)^1/2 d3x/2
=1/3*arcsin1.5x+C

Answer 5

∫1/[(4-9X^2)^1/2]dx
=1/3*∫(1-(3x/2)^2)d(3x/2)
=1/3*arcsin(3x/2)+C