Problem 276 - Calculus Volume 1

Question

In the following exercises, use a suitable change of variables to determine the indefinite integral. $$ \int \sin ^{7} \theta \cos \theta d \theta $$

Step-by-Step Solution

Answer

The integral is \( \frac{(\sin \theta)^8}{8} + C \).

Step 1: Identify the Substitution

Recognize that a direct substitution can simplify the integration process. Here, notice that the derivative of \( \sin \theta \) is \( \cos \theta \). This suggests we use a substitution of \( u = \sin \theta \), then \( du = \cos \theta d\theta \).

Step 2: Rewrite the Integral

Substitute \( u = \sin \theta \) and \( du = \cos \theta d\theta \) into the integral. The integral becomes: \( \int u^7 \, du \).

Step 3: Integrate with Respect to New Variable

Integrate \( \int u^7 \, du \). This is a standard power integral, which gives: \[ \frac{u^8}{8} + C \].

Step 4: Substitute Back to Original Variable

Replace \( u \) with \( \sin \theta \) from our substitution: \[ \frac{(\sin \theta)^8}{8} + C \]. This represents the solution in terms of the original variable \( \theta \).

Key Concepts

Integration techniques

Integration techniques are essential strategies used in calculus to solve integrals. Understanding these techniques can simplify complex problems and make the integration process more manageable.
Here are some common integration techniques you might encounter:

• Substitution: This technique involves changing the variable of integration to simplify the function. It's particularly useful when you recognize part of the integral that is the derivative of another function.
• Integration by Parts: Based on the product rule for differentiation, this method is handy for integrals involving products of functions.
• Partial Fraction Decomposition: Used for breaking down complex rational functions into simpler fractions to integrate more easily.
• Trigonometric Integration: Involves special trigonometric integrals and identities to tackle integrals with trigonometric functions.

Using the right technique depends on the form of the function you are integrating. Often, recognizing patterns or familiar derivatives is key to choosing the optimal method.
This exercise specifically utilizes the substitution method, which simplifies the integral by reducing the function to a basic power integral.

Indefinite integrals

In calculus, indefinite integrals are a type of integral that represent a family of functions. They lack specified limits of integration, unlike definite integrals.
An indefinite integral is represented as:\[\int f(x) \, dx = F(x) + C\]This expression means that integration results in a function \( F(x) \), plus a constant \( C \). This constant represents any number that could have been added to the original function before differentiation.

• The primary goal of finding an indefinite integral is to determine the antiderivative of a function.
• Indefinite integrals are crucial when analyzing relationships or solving differential equations.

They allow mathematicians to understand how quantities accumulate and change over time. In the exercise given, we're tasked with determining the indefinite integral of a trigonometric function modified using the substitution method.

Substitution method

The substitution method is a powerful integration technique that simplifies the integration process by introducing a new variable. This method is akin to the chain rule in reverse.
Here's a step-by-step breakdown of using the substitution method:

• Identify Substitution: Look for a function and its derivative within the integral that can be substituted to simplify the expression. In our problem, recognizing \( \sin \theta \) and its derivative \( \cos \theta \) guided our substitution choice.
• Rewrite the Integral: Substitute the new variable into the integral and change the variable of integration accordingly. This transforms the integral into a simpler format, often a standard power integral.
• Integrate: Solve the simpler integral using basic integral rules.
• Reverse the Substitution: Substitute back the original variable to express the result in terms of the initial variable.

In our exercise, we used \( u = \sin \theta \) and \( du = \cos \theta d\theta \), transforming the integral into \( \int u^7 \, du \).
This simplifies the function into a simple polynomial form, easy to integrate, and ultimately revert back to the original variable \( \theta \).