Integration

To solve the integral , we need to employ a trigonometric identity to simplify the integrand.

The function can be rewritten using the cosecant function: .

Recall the identity for cosecant:

Now, let's consider the identity for cosecant and its antiderivative:

• The integral of is:

Using substitution, set which implies .

The integral becomes:

Using the known antiderivative result:

This integral does not match any of the given options, confirming that the correct answer is "None of the above."

To find the centre of the circle that passes through and and touches the circle , we will solve step-by-step.

1. The given circle has its center at and radius .
2. Let the centre of the required circle be and its radius be .
3. Since the required circle passes through points and , these points satisfy the circle equation:
   and .
4. Since this circle is tangent to the circle , the distance between the centers and is: Thus, .
5. From steps 3 and 4, we have two equations:
   Equating the two expressions for gives:
6. Simplify the above equation: Cancel from both sides:
7. Solving for , we get: Since the radius cannot be negative, reducing further, and testing with options, yields a center located such that the effective distance leads to .
8. The only value for that allows real solution at location places the circle correctly orientated allowed by conditions.

Thus, the center of the circle is .