Problem 40 - Applied Calculus

Question

Find the integrals .Check your answers by differentiation. $$\int \frac{q}{5 q^{2}+8} d q$$

Step-by-Step Solution

Answer

The integral is \(\frac{1}{10} \ln |5q^2 + 8| + C\).

Step 1: Identify the Integral and Suggest Substitution

The given integral is \( \int \frac{q}{5q^2 + 8} dq \). In problems like this, where the integrand is a rational function, a common approach is to use substitution. Notice that the denominator is a function of \(q\) and its derivative appears in the numerator. Let's suggest \(u = 5q^2 + 8\).

Step 2: Differentiate and Express dq in Terms of du

Differentiate \(u = 5q^2 + 8\) to find \(du/dq\). We have \(du/dq = 10q\). Thus, \(du = 10q \, dq\). We need \(q \, dq\) in the integral, so solve for \(q \, dq\): \(q \, dq = \frac{du}{10}\).

Step 3: Substitute and Simplify Integral

Substitute \(u = 5q^2 + 8\) and \(q \, dq = \frac{du}{10}\) into the integral. The integral becomes: \( \frac{1}{10} \int \frac{1}{u} \, du \).

Step 4: Integrate with Respect to u

This is a straightforward integral: \( \frac{1}{10} \int \frac{1}{u} \, du = \frac{1}{10} \ln |u| + C\), where \(C\) is the integration constant.

Step 5: Substitute Back to Original Variable

Replace \(u\) with the original expression in terms of \(q\): \( \frac{1}{10} \ln |5q^2 + 8| + C\).

Step 6: Verify by Differentiation

Differentiate the result \( \frac{1}{10} \ln |5q^2 + 8| + C\) with respect to \(q\). Using the chain rule, \( \frac{d}{dq}[\frac{1}{10} \ln |5q^2 + 8| + C] = \frac{1}{10} \cdot \frac{1}{5q^2+8} \cdot 10q = \frac{q}{5q^2+8}\), which matches the original integrand. This confirms our integration is correct.

Key Concepts

Substitution Method

When faced with the task of integrating certain functions, especially rational ones, the substitution method is a handy tool. It's like solving a puzzle by changing the pieces to make it easier.
In this approach, you make a substitution to simplify the problem. For the given integral \( \int \frac{q}{5q^2 + 8} \, dq \), we choose a substitution that simplifies the expression inside the integral.

• Identify a function inside the integral whose derivative is also present. Here, the denominator \(5q^2 + 8\) has its derivative, \(10q\), in the numerator.
• Set \(u = 5q^2 + 8\), and compute its differential \(du = 10q \, dq\). This transformation shifts the focus to \(u\), making it easier to integrate.

By substitution, you reframe the integral in terms of \(u\), drastically simplifying what may initially seem complicated.

Differentiation

Once you find the integral using a method like substitution, you can verify your work by differentiating. This is a step that serves as a check to ensure that your integration was done correctly, effectively closing the loop.
Differentiation is the process of finding the derivative of a function. It is, essentially, the reverse of integration. For our solution, after substituting back to the original variable, the function was \( \frac{1}{10} \ln |5q^2 + 8| + C \).

• You differentiate this expression with respect to \(q\).
• Using the chain rule, the derivative of \( \ln |5q^2 + 8| \) involves multiplying by the derivative of \(5q^2 + 8\), which is \(10q\).

The derivative \( \frac{q}{5q^2 + 8} \) matches our original integrand, confirming the accuracy of our solution.

Rational Function Integration

Rational function integration typically involves expressions like \( \int \frac{P(q)}{Q(q)} \, dq \), where both \(P(q)\) and \(Q(q)\) are polynomials. In this realm, finding an integral can often seem complex, but with methods like substitution, things become more approachable.The integral \( \int \frac{q}{5q^2 + 8} \, dq \) is a type of rational function where the degree of the polynomial in the numerator is less than the degree in the denominator. This is where substitution thrives.

• Understand that the goal is to transform the integral into a basic form that is easily solvable, such as \( \int \frac{1}{u} \, du \).
• Simplification via substitution leads to a natural logarithmic integration, a common outcome when dealing with rational functions.

Finally, rational function integration becomes less daunting once you practice converting tricky parts of the integral into something you can easily manage and integrate.