Problem 10 - Thomas Calculus

Question

Evaluate the integrals. \(\int 4 x \sec ^{2} 2 x d x\)

Step-by-Step Solution

Answer

The integral evaluates to \( 2x \tan 2x - \ln |\sec 2x| + C \).

Step 1: Identify the Integral

The integral we need to evaluate is \( \int 4x \sec^2 2x \, dx \). This is a product of a polynomial term \(4x\) and a trigonometric function \(\sec^2 2x\).

Step 2: Apply Integration by Parts Formula

Integration by parts is given by the formula \( \int u \, dv = uv - \int v \, du \). Here, we choose \( u = 4x \) and \( dv = \sec^2 2x \, dx \).

Step 3: Differentiate and Integrate Parts

Differentiate \( u \) to get \( du = 4 \, dx \). Integrate \( dv \) to get \( v = \frac{1}{2} \tan 2x \), since \( \int \sec^2 2x \, dx = \frac{1}{2} \tan 2x \).

Step 4: Plug into Integration by Parts Formula

Substitute into the integration by parts formula: \( \int 4x \sec^2 2x \, dx = 4x \cdot \frac{1}{2} \tan 2x - \int \frac{1}{2} \tan 2x \cdot 4 \, dx \). This simplifies to: \( 2x \tan 2x - 2 \int \tan 2x \, dx \).

Step 5: Solve the Remaining Integral

The remaining integral is \( 2 \int \tan 2x \, dx \). The integral \( \int \tan 2x \, dx = \frac{1}{2} \ln |\sec 2x| + C \). So, multiplying by 2 gives us \( \ln |\sec 2x| + C \).

Step 6: Combine the Results

Substitute back into the equation to get the final result: \( 2x \tan 2x - \ln |\sec 2x| + C \).

Step 7: Final Answer

The evaluated integral is \( 2x \tan 2x - \ln |\sec 2x| + C \).

Key Concepts

Trigonometric Integration

Trigonometric integration is used when dealing with integrals involving trigonometric functions. These integrals are common in calculus and can often be solved using different techniques such as substitution or by parts. In the context of the exercise, we have a trigonometric function \( \sec^2 2x \) which is integrated alongside a polynomial, making it suitable for integration by parts.

Key trigonometric integrals to remember include:

• \( \int \sin x \, dx = -\cos x + C \)
• \( \int \cos x \, dx = \sin x + C \)
• \( \int \sec^2 x \, dx = \tan x + C \)

For the given integral, recognizing that \( \int \sec^2 x \, dx = \tan x + C \) helps us find \( v \) in the integration by parts process.

This approach highlights how different integrals can be connected, and learning the foundation of trigonometric identities and integrals is crucial in solving more complex calculus problems.

Integral Calculus

Integral calculus is the branch of calculus concerned with the concept of integration. It is opposite to differentiation and involves finding the function, given its derivative. It is especially important for calculating areas under curves and finding antiderivatives.

In real-world applications, integral calculus helps to solve problems involving motion, areas, volumes, and other accumulations. It is built on two fundamental ideas:

• Indefinite Integrals: Finding a function whose derivative is the given function, often represented as \( \int f(x) \, dx = F(x) + C \).
• Definite Integrals: Calculating the area under a curve from one point to another, \( \int_{a}^{b} f(x) \, dx \).

In solving the provided integral \( \int 4x \sec^2 2x \, dx \), we effectively apply indefinite integral techniques, specifically integration by parts. This highlights how integral calculus provides tools for breaking down complex expressions into manageable pieces.

Functions Differentiation

Functions differentiation is the process of finding the derivative of a function, which represents the rate of change. In calculus, differentiation and integration are inverse processes. Differentiation breaks functions down into their instantaneous rate of change, while integration combines small pieces to form a whole.

Key rules of differentiation include:

• Power Rule: \( \frac{d}{dx} x^n = nx^{n-1} \)
• Product Rule: \( \frac{d}{dx} (uv) = u'v + uv' \)
• Chain Rule: \( \frac{d}{dx} f(g(x)) = f'(g(x))g'(x) \)

In our problem, differentiation was used to calculate \( du \) from \( u = 4x \), resulting in \( du = 4 \, dx \). This step is critical as it provides part of the information needed to apply the integration by parts formula, thus linking differentiation directly to the process of solving an integral.