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.
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.