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Understanding u-substitution is essential for calculating integrals that are not in a straightforward form. This technique involves a change of variable to simplify the integrand—the mathematical expression being integrated—and often to make an antiderivative easier to find. The process starts with selecting a substitution variable, often denoted as 'u', which is a function of the original variable of integration.
In our exercise, the substitution chosen is \(u = \frac{1}{x}\). Once the substitution is made, the differential of u with respect to x (usually written as \(du\)) is found by differentiating the substitution. In this case, that differentiation led to \(du = -\frac{1}{x^2}dx\). This step is crucial because it allows us to express the infinitely small change in x (\(dx\)) in terms of u, resulting in a new, often more manageable integral.
After substituting \(dx\), the next step is to rewrite the entire integral in terms of u. The original integral, now expressed as \(\int -\sin(u)\ du\), is simpler and more recognizable. This process illuminates the power of u-substitution: transforming an intimidating integral into one that is more amenable to standard integration techniques.

Calculating a definite integral involves finding the net area under a curve between two points on the x-axis, which are called the bounds of integration. This is a fundamental concept in calculus because it provides us with the precise value of the accumulation of quantities, such as area, volume, displacement, among others.
In the given exercise, the process starts with an integral that is not easily solvable. The application of u-substitution transforms it into one that is recognizable as having a straightforward antiderivative. The definite integral is then evaluated by taking the antiderivative of the integrand and computing its value at the upper and lower bounds. For our exercise, after the u-substitution, we found that the antiderivative of \(\sin u\) is \(\cos u\), making the evaluation of the integral a matter of simple substitution at the new bounds and finding the difference.

Trigonometric functions such as sine (sin), cosine (cos), and tangent (tan) are common in calculus problems, and knowing their antiderivatives is key to solving integrals that involve them. For example, the antiderivative of \(sin x\) is \(\cos x\), while the antiderivative of \(\cos x\) is \(\sin x\), and so on. The antiderivative is the function which, when differentiated, yields the function we started with.
In our problem, once the u-substitution is applied, the integrand becomes \(\sin u\), allowing us to use our knowledge of trigonometric antiderivatives to find that \(\cos u\) is the antiderivative we need for evaluating the integral. This step is a pivotal part of the calculation, as it turns the previously complex expression into one where the trigonometric identity we are familiar with can be applied directly.

When we apply the u-substitution method to a definite integral, we must remember to adjust the bounds of integration accordingly. This ensures that the area calculation is performed over the correct interval. For a variable substitution \(u=g(x)\), where \(x\) is the original variable and \(g(x)\) is the function we're substituting with, the new bounds are found by substituting the original bounds into the substitution equation.
In our sample exercise, the original lower bound was \(x = \frac{1}{\pi}\) and the upper bound was \(x = \frac{2}{\pi}\). After applying the substitution \(u = \frac{1}{x}\), we have the transformed lower bound as \(u = \pi\) and the upper bound as \(u = \frac{\pi}{2}\). This change of bounds is a critical step in the process; if neglected, the definite integral would yield an erroneous result. Always check that the bounds match the variable of integration, especially after a substitution, to ensure the final answer is valid.