Problem 23 - Applied Calculus

Question

Find the integrals in problems. Check your answers by differentiation. $$ \int \sin ^{6} \theta \cos \theta d \theta $$

Step-by-Step Solution

Answer

The integral is \( \frac{\sin^7 \theta}{7} + C \).

Step 1: Setup the Integration by Parts

To integrate \( \int \sin^6 \theta \cos \theta \, d\theta \), notice that it can be simplified by a substitution. Let \( u = \sin \theta \), then \( du = \cos \theta \, d\theta \). The integral becomes \( \int u^6 \, du \).

Step 2: Integrate the Substituted Expression

Now solve \( \int u^6 \, du \). Use the power rule for integration: \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \). Thus, \( \int u^6 \, du = \frac{u^7}{7} + C \).

Step 3: Substitute Back to the Original Variable

Substitute \( u = \sin \theta \) back into the solution: \( \frac{(\sin \theta)^7}{7} + C \) or \( \frac{\sin^7 \theta}{7} + C \).

Step 4: Check the Solution by Differentiation

Differentiate the result \( \frac{\sin^7 \theta}{7} + C \) to ensure it gives the original integrand. Use the chain rule: \( \frac{d}{d\theta} \left( \frac{\sin^7 \theta}{7} \right) = \sin^6 \theta \cdot \cos \theta \). This matches the integrand, confirming the solution is correct.

Key Concepts

Integration by Parts

Integration by parts is a useful technique for solving integrals where simple substitution is not enough. The idea is based on the product rule for differentiation.\ Let's remember the formula for integration by parts:

• \( \int u \, dv = uv - \int v \, du \)

Here, \( u \) and \( v \) are functions of the variable you're integrating with respect to.
To choose \( u \) and \( dv \) wisely, the acronym "LIATE" can help: Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, and Exponential functions, in order of preference for choosing \( u \).
In this exercise, integration by parts wasn't directly used, but understanding its components can broaden your integration toolkit.

Substitution Method

The substitution method is an excellent strategy when dealing with integrals involving complex expressions. The key idea is to transform the integral into a simpler one by a suitable substitution. Here's how it typically works:

• Choose a substitution \( u = g(x) \) that simplifies the integral.
• Replace \( dx \) with the derivative \( du = g'(x) \, dx \).
• Rewrite the integral in terms of \( u \).

In our exercise, we cleverly chose \( u = \sin \theta \), which directly simplified \( \int \sin^6 \theta \cos \theta \, d\theta \) into \( \int u^6 \, du \). This made the integration process straightforward.
Once integrated, it is equally important to substitute back the original variable, completing the process.
This method not only simplifies integrals but also teaches how to recognize patterns within functions.

Power Rule for Integration

The power rule is one of the fundamental integration techniques. It applies to any function of the form \( \int x^n \, dx \), where \( n \) is not equal to \(-1\). The power rule is expressed as:

• \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \)

This rule was the finishing key to solving the transformed integral in our problem: \( \int u^6 \, du \). Applying the rule, we found \( \frac{u^7}{7} + C \).
It is crucial in integration problems as it works on various polynomial expressions.
Don't forget: Always add \( C \), the constant of integration, as indefinite integrals represent families of functions differing by a constant.