Problem 20 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

Find the indicated integral. $$\int \cos (1-2 t) d t \quad\( \)

Step-by-Step Solution

Answer

-\frac{1}{2} \sin(1 - 2t) + C

Step 1: Identify the integral and make a substitution

To solve the integral \(\int \cos (1-2 t) d t\), let’s make a substitution to simplify it. Choose \(u = 1 - 2t\).

Step 2: Compute the derivative of the substitution

Find the derivative of \(u\). If \(u = 1 - 2t\), then the derivative of \(u\) with respect to \(t\) is \(\frac{du}{dt} = -2\).

Step 3: Solve for dt

Rearrange \(\frac{du}{dt} = -2\) to solve for \(dt\). This gives \(dt = \frac{du}{-2}\).

Step 4: Substitute into the integral

Substitute \(u\) and \(dt\) into the integral: \(\int \cos (1-2 t) d t = \int \cos(u) \frac{du}{-2}\).

Step 5: Solve the new integral

Now integrate with respect to \(u\): \(\int \cos(u) \frac{du}{-2} = \frac{1}{-2} \int \cos(u) du\). The integral of \(\cos(u)\) is \(\sin(u)\), so the integral becomes \(\frac{1}{-2} \sin(u) + C\).

Step 6: Replace \(u\) with the original variable

Since \(u = 1 - 2t\), substitute back to get \(-\frac{1}{2} \sin(1 - 2t) + C\).

Key Concepts

Substitution Method

The substitution method is a powerful tool in calculus for solving integrals. It involves changing the variable of integration to simplify the integral.
Essentially, we substitute a part of the integral with a new variable.
This makes the integral easier to handle. In our case, we solved the integral of \( \int \cos (1-2 t) d t \) by substituting u = 1 - 2t.
Here's how it works step-by-step:

• Identify a part of the integrand to substitute
• Express this part as a new variable, say u
• Find the derivative of u with respect to the original variable
• Solve for the differential (dt in our case)
• Substitute back into the integral

It's a straightforward yet effective technique, especially when dealing with complex integrands.

Definite Integrals

Definite integrals compute the area under a curve between two points. This is different from an indefinite integral, which includes a constant of integration (C).
For a definite integral, we have limits of integration, say a and b.
When we perform the integration, we evaluate the antiderivative at these limits. Here's the basic idea:

• Find the antiderivative, F(x), of the function
• Evaluate F(x) at the upper limit and lower limit
• Subtract the value at the lower limit from the value at the upper limit

So, if we had limits in our original problem, we'd substitute these limits into \(-\frac{1}{2} sin(1-2t)\) after integrating, and perform the subtraction to get the final answer.

Trigonometric Functions

Trigonometric functions like sine and cosine frequently show up in calculus, especially in integrals.
They represent periodic functions, which are functions that repeat their values in regular intervals.
In our integral, we had the cosine function, \(\cos x\). Key points about trigonometric functions include:

• The integral of \(\cos(u) \) is \(\sin(u) + C \)
• The integral of \(\sin(u)\) is \(-\cos(u) + C\)

Remember, sine and cosine functions are derivatives of each other but with a sign change. These properties make solving integrals involving trigonometric functions systematic and straightforward if you know these basic integrals.