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Polar coordinates offer a powerful method to work with certain geometrical figures, like circles and disks, which can be cumbersome in rectangular coordinates. When utilizing polar coordinates, we express a point in the plane in terms of

• a radius, $$r$$, which measures the distance from the origin to the point, and
• an angle, $$\theta$$, which measures the direction of the point from the positive x-axis.

In the context of our problem, a disk centered at the origin with radius 2 is easily expressed with polar coordinates: the radius $$r$$ varies from 0 to 2 and the angle $$\theta$$ wraps around from 0 to $$2\pi$$, completing a full circle.

Furthermore, the transformation from rectangular to polar coordinates also modifies the area element $$dA$$. Instead of using $$dx\,dy$$, we use $$r\,dr\,d\theta$$, which accounts for the radial spread of the area measure outward from the origin.

Using polar coordinates simplifies solving integrals over circular regions, like transforming integrals of functions dependent on $$x^2 + y^2$$ into those involving $$r^2$$, hence making computations more manageable.

The substitution method is a vital technique in calculus, especially useful in integral calculations involving transformations for simplification. In the context of our integral,

• we focus on the substitution $$u = r^2$$, which alters the variable of integration from $$r$$ to $$u$$.
• This leads to a reformulation of the differential, where $$du = 2r\,dr$$ or equivalently $$r\,dr = \frac{1}{2}du$$.

Such a substitution makes the integral more straightforward to work with. The limits of integration also modify accordingly:

• when $$r = 0$$, $$u = 0$$,
• and when $$r = 2$$, $$u = 4$$.

This transforms our original inner integral over $$r$$ into an integral over $$u$$, simplifying the integration process by converting the integrand into a more recognizable form, such as $$\sin(u)$$.

These steps demonstrate the power of substitution: it can morph complex expressions into simpler forms, and thus help in evaluating integrals more effectively.

Trigonometric integrals are a particular type of integration involving functions like sine and cosine. These functions often appear in different forms, especially when dealing with polar coordinates. In our problem, the integrand is $$\sin(r^2)$$, which becomes after substitution, $$\sin(u)$$.

The integral of $$\sin(u)$$ is well-known and straightforward:

• it is evaluated as $$-\cos(u)$$, a fundamental antiderivative in calculus.

With defined limits from our substitution method, the integral becomes $$\frac{1}{2} \int_{0}^{4} \sin(u) \, du$$, which computes to:

• $$-\frac{1}{2} [\cos(u)]_{0}^{4} = -\frac{1}{2} (\cos(4) - \cos(0))$$.

Breaking it down, this results in evaluating the cosine function at the upper and lower bounds, where $$\cos(0)$$ equals 1. Thus, the integral resolves to

• $$-\frac{1}{2} (\cos(4) - 1)$$.

Trigonometric functions exhibit periodic behavior which can simplify the analytical steps in integral calculus, especially when integrated over regions like full circles in polar coordinates.