Problem 22 - Applied Calculus

Question

Find the integrals in problems. Check your answers by differentiation. $$ \int \frac{t}{1+3 t^{2}} d t $$

Step-by-Step Solution

Answer

\( \int \frac{t}{1+3t^2} \, dt = \frac{1}{6} \ln |1+3t^2| + C \). Verified by differentiation.

Step 1: Identify the Integral Form

The given integral is \( \int \frac{t}{1+3t^2} \, dt \). We need to find a function whose derivative is \( \frac{t}{1+3t^2} \).

Step 2: Choose a Suitable Substitution

To simplify the integral, choose the substitution \( u = 1 + 3t^2 \). This gives \( du = 6t \, dt \) or \( \frac{du}{6} = t \, dt \). Substituting these into the integral, it becomes \( \int \frac{1}{u} \cdot \frac{du}{6} \).

Step 3: Simplify the New Integral

The integral \( \int \frac{1}{u} \cdot \frac{du}{6} \) can be simplified to \( \frac{1}{6} \int \frac{1}{u} \, du \).

Step 4: Evaluate the Simplified Integral

The integral \( \int \frac{1}{u} \, du \) is \( \ln |u| + C \). Thus, \( \frac{1}{6} \int \frac{1}{u} \, du = \frac{1}{6} \ln |u| + C \).

Step 5: Substitute Back the Original Variable

Replace \( u \) with \( 1 + 3t^2 \). Thus, the integral becomes \( \frac{1}{6} \ln |1 + 3t^2| + C \).

Step 6: Differentiate to Check

Differentiate \( \frac{1}{6} \ln |1 + 3t^2| + C \) with respect to \( t \). Using the chain rule, the derivative is \( \frac{1}{6} \cdot \frac{1}{1+3t^2} \cdot 6t = \frac{t}{1+3t^2} \), which matches the original integrand.

Key Concepts

Substitution Method

The substitution method is a powerful technique used in integral calculus to simplify integration tasks. It involves changing the variable of integration to make the integral easier to solve. In this problem, we needed to find the integral of \( \int \frac{t}{1+3t^2} \, dt \). To do this, we chose a new variable \( u \) such that \( u = 1 + 3t^2 \). Doing so transforms the integrand into a simpler form.

Here's why it's effective:

• It converts a complex expression into an easily recognizable one.
• It allows us to use integration rules more straightforwardly.

In our example, once we set \( u = 1 + 3t^2 \), the differential \( du = 6t \, dt \) forms. Rewriting, we get \( \frac{du}{6} = t \, dt \). By substituting these into the integral, we simplify it to \( \frac{1}{6} \int \frac{1}{u} \, du \), which can then be easily integrated.

This method not only simplifies the problem but also provides a clear path to obtaining an answer. Once integrated, we must remember to substitute back using the original variable \( t \) to obtain the final solution.

Differentiation Check

After solving an integral, it's essential to verify the result. The differentiation check helps confirm that the solution is correct by differentiating the obtained result and ensuring it equals the original integrand.

In our solution, after finding that \( \int \frac{t}{1+3t^2} \, dt = \frac{1}{6} \ln |1 + 3t^2| + C \), we need to differentiate this expression to verify its correctness.

Using the chain rule, the derivative of \( \frac{1}{6} \ln |1 + 3t^2| + C \) with respect to \( t \) involves two steps:

• First, the derivative of the natural log function is \( \frac{1}{1 + 3t^2} \).
• Then, multiply by the derivative of the inside function \( 1 + 3t^2 \), which is \( 6t \).

After performing these steps, we find the derivative is \( \frac{t}{1+3t^2} \), which matches the original integrand. If everything checks out, it confirms that our integration was performed correctly.

Integral Evaluation

Evaluating integrals is the process of finding the antiderivative or the function whose derivative equals the integrand. Integral evaluation involves utilizing standard rules and methods to find this function.

In the provided exercise, after employing the substitution method, the task becomes simpler. We arrived at the integral \( \frac{1}{6} \int \frac{1}{u} \, du \). This is a well-known integral form:

• The integral of \( \int \frac{1}{u} \, du \) is \( \ln |u| + C \).
• So, the evaluated integral becomes \( \frac{1}{6} \ln |u| + C \).

The evaluated result shows the general form, important constant of integration \( C \), and emphasizes the method's efficiency.

Don't forget: After integration, substituting back with the original variable is crucial to express the answer in terms of the original variable \( t \). The final evaluated integral then is \( \frac{1}{6} \ln |1 + 3t^2| + C \), providing the solution we set out to find. Integral evaluation completes the task by giving a checkable solution.