`int_0^4 1/sqrt(25-x^2) dx` Evaluate the integral

The function under the integral is similar to the derivative of `arcsin(x),`   `1/sqrt(1 - x^2),` and may be reduced to it.

Namely, make the substitution `x = 5u,` then `dx = 5 du.` Also this means  `u = x/5`  and therefore the limits of integration for `u` are from `0/5 = 0` to  `4/5` (substitute the limits for `x` to the formula `u = x/5` ).

The integral becomes

`int_0^(4/5) (5 du)/sqrt(25 - 25u^2) =int_0^(4/5) (du)/sqrt(1 - u^2) =`

`= arcsin(4/5) - arcsin(0) = arcsin(4/5) approx 0.93.`