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U-substitution is an essential technique used in calculus to simplify the process of integrating more complex functions. This method involves changing variables to rewrite the integral in a simpler form. In essence, you choose a part of the integral to be the new variable, usually denoted as "u," that makes the rest of the function easier to work with.

Here's how it works:

• First, you identify a function inside the integral that, when differentiated, appears elsewhere in the integrand. In the given integral: \[\int_{1}^{2} \frac{1-\cos \theta}{\theta-\sin \theta} d \theta\]you choose the numerator \(1-\cos \theta\) as \(u\).
• Next, calculate the derivative of \(u\), which is \(\frac{du}{d\theta} = \sin \theta\). You need this to replace \(d\theta\) in the original integral.
• Rewrite the entire expression in terms of \(u\). This typically means getting a new expression for \(d\theta\) using \(du\) and substituting back into the integral.

After performing these steps, the complex integral transforms into something simpler, like the one we ended up with: \[ \int \frac{1}{u} du \]This new format is much easier to integrate.

A derivative represents the rate of change of a function with respect to a variable. It is a fundamental concept in calculus, widely used for solving integrals, as seen in u-substitution.

When choosing the substitution \(u = 1 - \cos \theta\), the derivative is key. Calculating the derivative provides:

• \( \frac{du}{d\theta} = \sin \theta \)

This derivative helps in expressing the differential of \(\theta\), \(d\theta\), in terms of \(du\). By knowing \(d\theta = \frac{du}{\sin \theta}\), it makes substitution possible within the integral. The derivative here acts like a bridge that connects the original variable \(\theta\) to \(u\).

Thus, finding and understanding derivatives is crucial in converting integrals into simpler, more manageable calculations through techniques like u-substitution. The entire integration becomes more straightforward when the function and its derivative are neatly aligned.

The natural logarithm, often denoted as \(ln(x)\), is a logarithm with a base \(e\) (where \(e \approx 2.718\)). It is widely used in integration, especially when dealing with the integral of the form \(\int \frac{1}{u} du\).

For our simplified integral result:\[ \int \frac{1}{u} du = ln|u| + C \]the function \(ln|u|\) arises naturally. This expression appears as a standard result when integrating \(\frac{1}{u}\) in calculus.

Once you've calculated the indefinite integral, substitute back the original expression for \(u\), in our case \(u = 1 - \cos \theta\). This places the natural logarithm within the context of the initial problem, leading to:\[ ln|1- \cos \theta| \]Understanding how the natural logarithm interacts with \(u\) in this way allows you to evaluate definite integrals effectively.

The evaluation of an integral, particularly a definite integral, involves calculating the area under a curve between two bounds. After substituting back the expression for \(u\) into the integrated result, our next step involves evaluating it between \(\theta = 1\) and \(\theta = 2\).

Here’s a concise step-by-step on how to do it:

• First, plug the upper limit into your expression: \(ln|1 - \cos 2|\).
• Next, substitute the lower limit: \(ln|1 - \cos 1|\).
• Finally, compute the difference between these two results: \[ ln|1- \cos 2| - ln|1- \cos 1| \]This expression determines the area under the curve from \(\theta = 1\) to \(\theta = 2\).

This process completes the integration solution by accounting for the specific limits given in the problem, thus evaluating the definite integral.