How should I approach this integral

evaluate the integral x^2 / (sqrt(64x^2 - 16)) dx for x > 1/2

My first thought was to set it up as an improper integral from (1/2)+ to infinity and then use

lim as c -> (1/2)+ of the integral of f(x) from c to pi/2 + lim d -> infinity of the integral of f(x) from pi/2 to d

i chose pi/2 because i then used trig substitution to solve the integral

i chose x = 1/2sin(t) dx = 1/2cos(t)

is this the correct way to do this problem the original question is exactly what the assignment asks. also i got stuck later on doing this way because of having to change all of the bounds. so if you can solve and let me know what you get, that would help as well

You are correct that you need to set this up as a proper integral; however, your substitution is incorrect.

To integrate expressions of the form √(x^2 - a^2), you need the substitution:
x = a*sec(u),

to exploit the fact that sec^2(u) - 1 = tan^2(u).

Here, substitute x = (1/2)sec(u) ==> dx = (1/2)sec(u)tan(u). The new bounds will be:
(a) Lower bound: x --> 1/2 ==> (1/2)sec(u) --> 1/2 ==> sec(u) --> 1 ==> u --> π/4
(b) Upper Bound: x --> infinity ==> u --> π/2.

The integral now becomes:
∫ x^2/√(64x^2 - 16) dx (from x=1/2 to infinity)
= lim (a, b)-->(1/2, infinity) ∫ x^2/√(64x^2 - 16) dx (from x=a to b)
= lim (a, b)-->(π/4, π/2) ∫ [(1/4)(1/2)sec^3(u)tan(u)]/√[(64)(1/4)se… - 16] du (from u=a to b)
= lim (a, b)-->(π/4, π/2) ∫ [(1/8)sec^3(u)tan(u)]/√[16sec^2(u) - 16] du (from u=a to b)
= lim (a, b)-->(π/4, π/2) ∫ [(1/8)sec^3(u)tan(u)]/[4sec(u)] du (from u=a to b)
= 1/32 * lim (a, b)-->(π/4, π/2) ∫ sec^2(u)tan(u) du (from u=a to b)
= 1/64 * lim (a, b)-->(π/4, π/2) [tan^2(u) (evaluated from u=a to b)]
= 1/64 * lim (a, b)-->(π/4, π/2) [tan^2(b) - tan^2(a)].

This limit is divergent since tan^2(b) --> infinity as b --> π/2, so the integral is divergent.

I hope this helps!