Problem 26 - Applied Calculus

Question

Find the integrals in problems. Check your answers by differentiation. $$ \int \sin ^{3} \alpha \cos \alpha d \alpha $$

Step-by-Step Solution

Answer

The integral is \( \frac{(\sin \alpha)^4}{4} + C \).

Step 1: Identify the Integrand Type and Substitution

Upon looking at the integrand \( \sin^3 \alpha \cos \alpha \), we notice that it can be simplified using a substitution method. Let \( u = \sin \alpha \). Then, \( du = \cos \alpha \, d\alpha \). This substitution transforms the integral into \( \int u^3 \, du \).

Step 2: Perform the Integration

Now, integrate \( \int u^3 \, du \) using the power rule for integration. The power rule states that \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \). Here \( n = 3 \), so the integral becomes:\[ \int u^3 \, du = \frac{u^4}{4} + C \].

Step 3: Reverse Substitution

Substitute back \( u = \sin \alpha \) into the integrated function. We have:\[ \frac{(\sin \alpha)^4}{4} + C \].

Step 4: Differentiate to Check

To confirm our integration, we differentiate \( \frac{(\sin \alpha)^4}{4} + C \) with respect to \( \alpha \).Using the chain rule, let \( v = \sin \alpha \), then \( \frac{dv}{d\alpha} = \cos \alpha \), and\( \frac{d}{d\alpha} \left( \frac{v^4}{4} \right) = v^3 \cdot \frac{dv}{d\alpha} = \sin^3 \alpha \cdot \cos \alpha \),which matches the original integrand.

Key Concepts

Substitution Method

The substitution method is a technique used to simplify integrals, making them easier to solve. The idea is to replace a more complicated expression with a simpler one.

Let's take our example integrand, \( \sin^3 \alpha \cos \alpha \), as a case study. Here, the expression is difficult to integrate directly, so we use the substitution method. We set \( u = \sin \alpha \) and find \( du = \cos \alpha \, d\alpha \).

• By substituting \( u \) for \( \sin \alpha \) and \( du \) for \( \cos \alpha \, d\alpha \), the integral \( \int \sin^3 \alpha \cos \alpha \, d\alpha \) simplifies to \( \int u^3 \, du \).
• This method transforms a complex trigonometric integral into a simple polynomial one.

So, the substitution method is all about making the problem easier to manage by "substituting" complicated parts with simpler variables.

Power Rule

The power rule is a fundamental rule in calculus used to integrate functions of the form \( x^n \). Understanding the power rule is essential for integrating polynomial expressions.

For any function \( \int x^n \, dx \), the power rule tells us that the integral is given by:\[\int x^n \, dx = \frac{x^{n+1}}{n+1} + C,\]where \( C \) is the constant of integration. It's important to note that \( n eq -1 \) because the power rule does not apply to the integral of \( x^{-1} \).

• In our worked example, once we applied the substitution \( u = \sin \alpha \), our integral became \( \int u^3 \, du \).
• Using the power rule, this simplifies to \( \frac{u^4}{4} + C \).

The power rule is thus a straightforward technique you can use to find the integral of any polynomial function of the form \( x^n \).

Chain Rule

While the chain rule is primarily a differentiation technique, it assists in confirming our integration results by checking our work. The chain rule states that if you have a function \( y \) that depends on \( u \), and \( u \) itself is a function of \( x \), then:\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}. \]

This rule is applied during the "check" process in integration. Let's see how it was used in our example:

• After finding the integral \( \frac{(\sin \alpha)^4}{4} + C \), we used the chain rule to differentiate it with respect to \( \alpha \).
• By letting \( v = \sin \alpha \), we differentiate \( \frac{v^4}{4} \) through \( \frac{dy}{dv} = v^3 \) and multiplying by \( \frac{dv}{d\alpha} = \cos \alpha \).

By doing so, we end up with the original integrand \( \sin^3 \alpha \cos \alpha \), confirming that our integration was done correctly.
The chain rule thus serves as a valuable tool for ensuring the accuracy of our integration by differentiation.