Problem 4 - Calculus: One and Several Variables

Question

Calculate. (If you run out of ideas, use the examples as models.) $$\int \cos ^{3} x d x$$.

Step-by-Step Solution

Answer

The antiderivative of \(\int \cos^3 x dx\) is \[ \sin x - \frac{1}{3} \sin^3 x + C. \]

Step 1: Use the reduction formula for cosine powers

We start by using the reduction formula for odd powers of cosine: \[ \cos^{2n+1} x = (1-\sin^2 x)^n \cos x. \] In our case, n = 1, so we have: \[ \cos^3 x = (1-\sin^2 x) \cos x. \] Now, our integral becomes: \[ \int \cos^3 x dx = \int (1-\sin^2 x) \cos x dx. \]

Step 2: Use substitution

Let \(u = \sin x\), so \(du = \cos x dx\). Now, our integral becomes: \[ \int (1-u^2) du. \]

Step 3: Integrate with respect to u

We can now compute the antiderivative with respect to u: \[ \int (1-u^2) du = u - \frac{1}{3} u^3 + C, \quad (C: constant) \]

Step 4: Replace u with sin x

Substitute back \(u = \sin x\) and get the final answer: \[ u - \frac{1}{3} u^3 + C = \sin x - \frac{1}{3} \sin^3 x + C. \] Thus, the antiderivative of \(\int \cos^3 x dx\) is \[ \sin x - \frac{1}{3} \sin^3 x + C. \]

Key Concepts

Reduction Formula

The reduction formula is an invaluable tool when dealing with integrals involving powers of trigonometric functions. It simplifies the integration process by **reducing** the power of the function step-by-step. For instance, when faced with an odd power of cosine, such as \( \cos^3 x \), the reduction formula rearranges it into a more manageable form:

• \( \cos^{2n+1} x = (1 - \sin^2 x)^n \cos x \)

In our example, we have \( n = 1 \). The reduction formula transforms \( \cos^3 x \) into:

• \( \cos^3 x = (1 - \sin^2 x) \cos x \)

This transformation allows us to set up the integral in a form that is ready for the substitution method, making the calculation much simpler. It's like setting the stage for the next act in our integration process.

Substitution Method

The substitution method is akin to a powerful magic trick in the world of integration. It works by transforming a complex integral into a simpler one, usually involving a single variable. Once you master this method, integrals that appeared daunting become straightforward. Let's break down the steps:

• Substitute a part of the integral with a new variable \( u \), turning the expression into something easier to handle.
• Find the differential (\( du \)) of the new function in terms of the original variable.

In our case, we set \( u = \sin x \), which transforms \( du \) to \( \cos x \, dx \). The integral \( \int (1 - \sin^2 x) \cos x \, dx \) now becomes:

• \( \int (1 - u^2) \, du \)

This simplification allows us to focus on a basic polynomial integral. By using substitution, we've shifted our problem into a much more digestible form.

Antiderivative

The concept of an antiderivative is central to the process of integration. Finding the antiderivative means determining the original function whose derivative yields the given function. In simpler terms, it's like going backwards in differentiation.For the integral \( \int (1-u^2) \, du \), the antiderivative is computed by integrating each term:

• For \( 1 \), we get \( u \).
• For \( -u^2 \), we get \( -\frac{1}{3}u^3 \).

Combining these results, the antiderivative is:

• \( u - \frac{1}{3} u^3 + C \)

Here, \( C \) is the constant of integration, always added when finding indefinite integrals. Finally, we return to the original variable by substituting back \( u = \sin x \), providing the solution:

• \( \sin x - \frac{1}{3} \sin^3 x + C \)

Thus, through careful steps, we've unveiled the antiderivative, highlighting the beautiful connection between differentiation and integration.