To integrate ∫ 2 dx / sin^2(x-5), we can start by realizing that sin^2(x) can be written as 1 - cos^2(x) using the Pythagorean identity.

Our given integral becomes:
∫ 2 dx / (1 - cos^2(x-5))

Let u = x - 5, then du = dx

Substitute u back into the integral:
∫ 2 du / (1 - cos^2(u))

Using the identity cos^2(u) = 1 - sin^2(u), we get:
∫ 2 du / sin^2(u)

Now, we can rewrite sin^2(u) as 1 - cos^2(u) and use the substitution v = cos(u) and dv = -sin(u) du:

∫ -2 dv / (1 - v^2)

This is a standard integral that can be solved using partial fractions or by recognizing that it integrates to -tanh^(-1)(v) and plugging back in v = cos(u):

-2 * tan^(-1)(cos(u)) + C

Substitute u = x - 5 back in:
-2 * tan^(-1)(cos(x-5)) + C

So, the integral of 2 dx / sin^2(x-5) is equal to -2 * tan^(-1)(cos(x-5)) + C.