`int_0^1(2\ dx)/(sqrt(9-4x^2`

Let `u=2x` then `du=2\ dx`.

Our integral becomes:

`int_0^1(2\ dx)/(sqrt(9-4x^2))`

`=int_(x=0)^(x=1)(du)/(sqrt((3)^2-u^2))`

`=[sin^-1u/3]_(x=0)^(x=1)`

`=[sin^-1 (2x)/3]_0^1`

`=[sin^-1 2/3-sin^-1 0]`

`=0.7297`

We could have done it in much fewer steps by leaving it in terms of "`u`", as follows:

`int_0^1(2 dx)/(sqrt(9-4x^2))`

`=int_0^2(du)/(sqrt((3)^2-u^2))`

`=[sin^-1 2/3-sin^-1 0]`

`=0.7297`

Note the change in the limits when `dx` is changed to `du` during the integration.

This is because we let `u = 2x`.

Hence the limits for `x` of

`x = 0` and `x = 1`

have to be changed for `u` to

`u = 0` and `u = 2`.