Problem 24 - Applied Calculus

Question

Find the integrals. Check your answers by differentiation. \(\int \sin \theta(\cos \theta \quad+\) 5)\(^{7} d \theta\)

Step-by-Step Solution

Answer

The integral is \(-\frac{(\cos \theta + 5)^8}{8} + C\). Verified by differentiation.

Step 1: Identify the Formula for Substitution

We begin by noticing that the integral has a composite function \( (\cos \theta + 5)^7 \). This suggests that substitution might be useful. We can use \( u = \cos \theta + 5 \) because the derivative of \( \cos \theta \) is \(-\sin \theta \), which appears in the integral.

Step 2: Substitute and Simplify

With \( u = \cos \theta + 5 \), we differentiate to find \( du = -\sin \theta \, d\theta \). Solving for \( \sin \theta \, d\theta \), we have \( \sin \theta \, d\theta = -du \). Substitute into the integral to get: \( \int (u)^7 (-du) \). This simplifies to \( -\int u^7 \, du \).

Step 3: Integrate Using the Power Rule

Apply the power rule for integrals: \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \). Thus, \( -\int u^7 \, du = -\frac{u^8}{8} + C \).

Step 4: Substitute Back to Original Variable

Return to the original variable \( \theta \) by substituting \( u = \cos \theta + 5 \) back into the expression: \( -\frac{(\cos \theta + 5)^8}{8} + C \).

Step 5: Differentiate to Verify

Differentiate \( -\frac{(\cos \theta + 5)^8}{8} + C \) with respect to \( \theta \):1. Use the chain rule: Derivative of outer function \( -\frac{1}{8} (fn)^8 \) is \( -\frac{8}{8}(\cos \theta + 5)^7 \times d(fn)/d\theta \).2. Multiply by \(-\sin \theta\), the derivative of \(\cos \theta\).3. Simplifying gives \( \sin \theta (\cos \theta + 5)^7 \), matching the original integrand.

Key Concepts

Substitution Method

The substitution method is an essential tool in calculus for solving integrals involving composite functions. Think of it as a change of variables. This process makes the integral simpler to evaluate.
To do this effectively, we identify a part of the integral that might be replaced with a single variable, say, \( u \). This variable is chosen because its derivative is also present in the integrand, making the substitution process seamless.

• Identify the inner function: Look for a composite function. In our example, \( \cos \theta + 5 \) is a good candidate.
• Define \( u \): Let \( u = \cos \theta + 5 \).
• Differentiate \( u \): Find \( du \), which here is \(-\sin \theta \, d\theta \).
• Substitute: Replace all occurrences in the original integral with \( u \) and \( du \).

By converting the integrand to a simpler form, you can apply other integration techniques more easily, bringing clarity and simplicity to the solution.

Power Rule for Integration

The power rule for integration is akin to the reverse of basic differentiation rules. It is a straightforward method to integrate functions of the form \( x^n \). Given an integral of \( u^n \), the power rule tells us how to determine its integral. The power rule for integration is written as:
\[\int u^n \, du = \frac{u^{n+1}}{n+1} + C,\]where \( u \) is the function being integrated, \( n eq -1 \), and \( C \) is the constant of integration.
Here's how you apply it:

• Increase the exponent by one: Transform \( u^n \) to \( u^{n+1} \).
• Divide by the new exponent: Multiply by \( \frac{1}{n+1} \).
• Add the integration constant \( C \): Always include \( C \) to represent any constant indefinite integral.

This rule efficiently handles polynomial expressions and is fundamental to integrating a wide array of functions.

Chain Rule Differentiation

The chain rule is a fundamental principle of calculus used to differentiate composite functions. Essentially, it allows us to handle the "outside" and "inside" functions separately when differentiating. Let's break down how it works:
Consider a function like \( f(g(x)) \). Here, \( g(x) \) is the inner function, and \( f \) is the outer function. The chain rule states:
\[\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x),\]where \( f' \) is the derivative of the outer function, and \( g' \) is the derivative of the inner function.

• Differentiate the outer function: Use the power (or other applicable) rule on \( f \).
• Multiply by the derivative of the inner function: Calculate \( g'(x) \) and multiply it with the result from the previous step.

In our solution verification, the chain rule helps us differentiate back to the original integrand, validating the integral you found. It is vital for confirming and connecting calculus concepts intuitively.