Verified answer

There is a general form

∫(a^2-(bx)^2)^-1/2 dx = (1/b)sin^-1 [bx/a] +C

∫(4-9x^2)^-1/2 dx = (1/3)(sin^-1 [3x/2])/3 +C

which is A

Answer: A

Look in the end papers of any calculus book to find the following integration formula ( I am using INT in place of the integral sign):

INT (a^2 - u^2)^(-1 / 2) du = arcsin (u / a) + C

So, rewrite INT (4 - 9x^2)^(-1 / 2) dx as

INT (2^2 - (3x)^2)^(-1 / 2) dx

Let a = 2 and u = 3x and du = 3 dx and (1 / 3) du = dx.

By substitution we have

INT ((1 / 3)(a^2 - u^2)^(-1 / 2) du

=(1 / 3) arcsin (u / a) + C

=(1 / 3) arcsin (3x / 2) + C

∫ (4 - 9x^2)^-1/2 dx =?

let sin y = (3x/2)

cos y dy = 3/2 dx === > dx = 2/3 cos y dy

(4 - 9x^2) = 4(1- sin^2y) = 4 cos^2 y

Int = ∫ [(2/3 cos y) / 2 (cos y)] dy

Int = ∫ 1/3 dy = 1/3 y +c = 1/3 sin^-1 (3x/2) + C

The correct answer is a.