Problem 47 - Thomas Calculus in SI Units

Question

Evaluate the integrals. $$\int \frac{3 \sec ^{2} t}{6+3 \tan t} d t$$

Step-by-Step Solution

Answer

The integral evaluates to \( \ln |2 + \tan t| + C \).

Step 1: Identify the Substitution

The integral involves a function of \( \tan t \). This suggests using substitution to simplify the expression. Let \( u = \tan t \). Then, the derivative \( du = \sec^2 t \, dt \).

Step 2: Substitute and Simplify the Integral

After substitution, change the variables in the integral: \[ \int \frac{3 \sec^2 t}{6 + 3 \tan t} \, dt = \int \frac{3}{6 + 3u} \, du \] Simplify the expression inside the integral: \[ \int \frac{3}{3(2 + u)} \, du = \int \frac{1}{2 + u} \, du \]

Step 3: Integrate the Simplified Expression

Now integrate \( \frac{1}{2 + u} \). The integral of \( \frac{1}{a+x} \) is \( \ln|a+x| + C \). So, \[ \int \frac{1}{2 + u} \, du = \ln|2 + u| + C \]

Step 4: Back-Substitute to Original Variable

Change back to the original variable \( t \). Since \( u = \tan t \), substitute \( u \) back: \[ \ln|2 + u| + C = \ln|2 + \tan t| + C \]

Step 5: Simplify if Necessary

The solution integrates directly to obtain the final answer. No further simplification is necessary beyond expressing the result cleanly.

Key Concepts

Substitution Method

The substitution method is a powerful tool for simplifying integrals, especially when dealing with complex expressions. The basic idea is to transform a difficult integral into a simpler one using a suitable substitution. Here’s how you can think of it:

• Identify parts of the integrand that can be exchanged with a single variable. This often involves picking a function inside of another function, such as the one in our exercise with \( \tan t \).
• Define a new variable, say \( u \), that encapsulates this part. In our example, \( u = \tan t \).
• Replace all occurrences of this part and its differential in the integral. This is where you see the result of the substitution method: the integrand becomes simpler, turning a tough problem into an easy one.

This method is particularly useful as it reduces the complexity of integration by lowering the degrees of polynomials or converting trigonometric functions into simpler forms.

Integration Techniques

In calculus, mastering integration techniques is crucial for solving a wide variety of integrals. Each technique has its own situation where it shines. Common methods include:

• Substitution, which is like using the chain rule in reverse, helps simplify the integration of composite functions.
• Partial Fraction Decomposition, used for rational functions, makes breaking fractions down into simpler terms possible for straightforward integration.
• Integration by Parts, inspired by the product rule for differentiation, is useful for integrals of products of functions.
• Trigonometric Integrals, which involve using identities and known integrals of sine, cosine, and other trigonometric functions, often need substitution or clever algebraic manipulation.

The choice of technique depends on the integrand's form. For instance, a rational function like \( \frac{3}{6 + 3u} \) in the exercise can often be simplified by looking for easy substitutions, reducing it to a basic form before integration.

Trigonometric Integrals

Trigonometric integrals involve integrals of products or powers of trigonometric functions. These can appear daunting at first, but with the right approach, they become manageable. Here's how you can tackle them:

• Use trigonometric identities where necessary to simplify the integrand. Common identities include \( \sin^2 x + \cos^2 x = 1 \) and angle sum formulas.
• Apply substitutions to convert trigonometric forms into algebraic ones. Our exercise did this with \( \tan t \) and its derivative \( \sec^2 t \) which made the integration process more straightforward.
• Integrals like \( \int \sec^2 t \, dt \) are direct results of differentiating basic trigonometric functions and should be memorized for quick solving.

By leveraging these strategies, dealing with trigonometric integrals can transit from being a hurdle to a smooth process, allowing for efficient problem-solving.