Let x = 4 sec θ then dx = 4 sec θ tan θ dθ and
x² = 16 sec²θ

Simplifying the square root part:

Substituting dx = 4 sec θ tan θ dq, x² = 16 sec²x and into the given integral gives us: [taking indefinite case first]

Since we let x = 4 sec θ , we get

Using a triangle, we can also derive that:

and

Therefore, we can conclude that:

NOTE: We could have changed upper and lower limits for θ here, and there would be no need to convert our expression back in terms of x.