Problem 43 - Calculus: One and Several Variables

Question

Calculate. (If you run out of ideas, use the examples as models.) $$\int_{0}^{z / 6} \tan ^{2} 2 x d x$$.

Step-by-Step Solution

Answer

The short answer is: \(-\frac{1}{2}\ln|\cos(\frac{z}{3})|\).

Step 1: Express \(\tan^2(2x)\) in terms of \(\sin\) and \(\cos\)

Recall that \(\tan^2(2x) = \frac{\sin^2(2x)}{\cos^2(2x)}\). Thus, our integral becomes: \[\int_{0}^{z / 6} \frac{\sin^2(2x)}{\cos^2(2x)} dx.\]

Step 2: Find an antiderivative of the resulting function

To find the antiderivative of \(\frac{\sin^2(2x)}{\cos^2(2x)}\), we can use the substitution method: Let \(u = \cos(2x)\), so \(du = -2\sin(2x)\cos(2x)dx\). We can rewrite \(\sin^2(2x)dx\) in terms of \(u\): \[\sin^2(2x) dx = -\frac{1}{2}\cos(2x) du = -\frac{1}{2}u du.\] Now, rewrite the integral in terms of \(u\): \[\int_{0}^{z / 6} \frac{\sin^2(2x)}{\cos^2(2x)} dx = -\frac{1}{2} \int \frac{u du}{u^2}.\] Now we can easily find the antiderivative: \[-\frac{1}{2} \int \frac{u du}{u^2} = -\frac{1}{2} \int \frac{1}{u} du = -\frac{1}{2}\ln|u| + C.\] Now, we need to express our antiderivative in terms of \(x\): \[-\frac{1}{2}\ln|\cos(2x)| + C.\]

Step 3: Evaluate the antiderivative at the endpoints of the interval, and subtract the results to find the definite integral

Now, we'll evaluate the integral \(-\frac{1}{2}\ln|\cos(2x)|\) at the endpoints \(\frac{z}{6}\) and \(0\): \[-\frac{1}{2}\ln|\cos(2(\frac{z}{6}))| - (-\frac{1}{2}\ln|\cos(2(0))|).\] ி Simplify: \[-\frac{1}{2}\ln|\cos(\frac{z}{3})| + \frac{1}{2}\ln|\cos(0)|.\] Since \(\cos(0) = 1\), the \(\ln|\cos(0)|\) term is equal to 0. Therefore, our final answer is: \[-\frac{1}{2}\ln|\cos(\frac{z}{3})|.\]

Key Concepts

Substitution Method

The substitution method is a powerful technique in calculus for simplifying integrals, especially when dealing with composite functions. It involves replacing a part of the integral with a new variable to make the integration process easier.

In the integral \(\int_{0}^{z / 6} \frac{\sin^2(2x)}{\cos^2(2x)} dx\), we used substitution to simplify the expression. We let \(u = \cos(2x)\), which leads to the differential \(du = -2\sin(2x)\cos(2x)dx\). This substitution transforms our integral into a new expression with respect to \(u\), namely \(-\frac{1}{2} \int \frac{u du}{u^2}\).

• Always identify a part of the function that can simplify the integrand when substituted.
• Ensure that the differential \(dx\) matches the derivative of your substitution.
• Change the limits of integration if performing a definite integral.

The goal is to rewrite the integral in terms of a new variable \(u\) such that integration becomes straightforward.

Trigonometric Integration

Trigonometric integration often requires the use of identities and transformations to simplify an integral. This is important with functions involving trigonometric expressions like \(\tan^2(2x)\).

Recall that \(\tan^2(2x) = \frac{\sin^2(2x)}{\cos^2(2x)}\). Understanding this relationship helps in simplifying or rewriting expressions in terms of basic trigonometric functions. By rewriting \(\tan\) in terms of \(\sin\) and \(\cos\), we can utilize various identities to find a path to easier integration.

• Use identities such as \(\sin^2(x) + \cos^2(x) = 1\) to rewrite functions.
• Recognize when a trigonometric substitution can simplify the problem.
• Direct computation is often simpler with these transformed expressions.

Effectively using trigonometric integration involves practice with identities and transformations to turn complex expressions into those that are more manageable.

Calculus Techniques

Calculus is equipped with a myriad of techniques to tackle integrals. In our example, a blend of these methods was necessary.

We see how the substitution method and trigonometric simplification come together. After finding the antiderivative \(-\frac{1}{2}\ln|\cos(2x)| + C\), calculus also instructs us on how to evaluate definite integrals by substitution into the antiderivative at the boundaries \([0, z/6]\).

• Always start by simplifying the integral if possible (e.g., using identities).
• Choose a suitable substitution to simplify the integration process.
• Evaluate the antiderivative at the boundaries for definite integration.

Mastery of these techniques requires understanding when and how to apply each effectively, dependent on the particular form of the function at hand. By practicing a range of problems, one becomes adept at quickly recognizing the best method to use.