Problem 286 - Calculus Volume 1

Question

In the following exercises, use a suitable change of variables to determine the indefinite integral. $$ \int(\cos \theta-1)\left(\cos ^{2} \theta-2 \cos \theta\right)^{3} \sin \theta d \theta $$

Step-by-Step Solution

Answer

Use substitution \(u = \cos \theta\), then integrate and simplify.

Step 1: Identify Substitution

We start by identifying a substitution that can simplify the integral. Notice the expression \((\cos^2 \theta - 2\cos \theta)^3\) is a function of \(\cos \theta\). Let's set \(u = \cos \theta\). Then, the derivative \(du = -\sin \theta\,d\theta\). We can rearrange this as \(-du = \sin \theta\,d\theta\).

Step 2: Simplify the Integral

Substitute \(u = \cos \theta\) into the integral. The integral becomes: \[\int (u - 1)(u^2 - 2u)^3 (-du)\]which can be rearranged as:\[\int -(u - 1)(u^2 - 2u)^3 \, du\]

Step 3: Distribute and Integrate

Distribution of the terms gives: \[-\int [u(u^2 - 2u)^3 - (u^2 - 2u)^3] \, du\].This can be written in two separate integrals:\[\int u(u^2 - 2u)^3 \, du - \int (u^2 - 2u)^3 \, du\].Evaluate each integral using standard integration methods.

Step 4: Evaluate the First Integral

For the integral \(\int u(u^2 - 2u)^3 \, du\), expand the polynomial \((u^2 - 2u)^3\) and then integrate term by term. This expansion results in a polynomial that can be integrated using the power rule of integration.

Step 5: Evaluate the Second Integral

For \(\int (u^2 - 2u)^3 \, du\), use the binomial expansion of \((u^2 - 2u)^3\) and integrate each term separately with respect to \(u\). Again, apply the power rule for each term obtained after expansion.

Step 6: Combine Results

Combine the results of each evaluated integral from Steps 4 and 5, simplifying where necessary to obtain the final result in terms of \(u\).

Step 7: Substitute Back \(\theta\)

Replace \(u\) with \(\cos \theta\) in your combined result from Step 6 to express the indefinite integral back in terms of \(\theta\).

Key Concepts

Change of Variables

When faced with a complicated integral, a common strategy is to use a technique called "change of variables" or "substitution." This approach is particularly useful when you have a composite function that becomes easier to work with after substituting a single variable. In the exercise given, the expression

• \((\cos^2 \theta - 2\cos \theta)^3\)
• is identified as a function of \(\cos \theta\).

To simplify the integration, we set \(u = \cos \theta\). This new variable \(u\) is much simpler to handle in terms of integration.

The key to remember is that when you replace \(\theta\) with \(u\), you also need to convert \(d\theta\) into \(du\). In this scenario, the derivative of \(\cos \theta\) is \(-\sin \theta\), leading to \(du = -\sin \theta\,d\theta\). Rearranging gives us \(-du = \sin \theta\,d\theta\). This small transformation is crucial for reducing the original integral into a simpler form in terms of \(u\). With this effective change of variables, integrative calculations often become much more straightforward.

Indefinite Integral

The problem at hand involves finding the indefinite integral of a specific expression. An indefinite integral, unlike a definite integral, does not have set limits. This kind of integral is essentially another name for the antiderivative, which represents a family of functions whose derivative is the integrand given. Inside the integral, our task is to simplify the expression by recognizing it as something more manageable. With the substitution \(u = \cos \theta\), the indefinite integral

• \(\int (\cos \theta - 1)(\cos^2 \theta - 2\cos \theta)^3 \sin \theta \,d\theta\)
• transforms into
• \(-\int (u - 1)(u^2 - 2u)^3 \, du\), which is simpler.

The goal of calculating the indefinite integral is to find out what function, when derived, leads us back to our original expression. This approach frequently employs the power rule of integration after expanding the polynomial in question. Each term in the polynomial can be dealt with separately, simplifying the problem of finding the antiderivative.

After working through the integration, don't forget to substitute back the original variable, reverting from \(u\) to \(\cos \theta\), thus completing the process and addressing the original variable context of the problem.

Trigonometric Substitution

Trigonometric substitution is a useful method for solving integrals that involve trigonometric functions. This method leverages the properties of trigonometric identities to simplify the integral. At first, you make a substitution with a trigonometric function that turns the integral into a more recognizable and easier form. One common scenario is when substituting for a trigonometric function like \(\cos \theta\) or \(\sin \theta\).

In the given integral, we started by setting \(u = \cos \theta\). This trigonometric substitution transforms the integral:

• \(\int (\cos \theta - 1)(\cos^2 \theta - 2\cos \theta)^3 \sin \theta \,d\theta\)
• into \(-\int (u - 1)(u^2 - 2u)^3 \, du\).

Such transformations are advantageous because working directly with polynomial and algebraic expressions is often easier than dealing with trigonometric ones. With the right substitution, integrals that initially seem complex due to the presence of trigonometric functions become much more approachable.

After the computation is complete, reversing the substitution reconnects us to the original trigonometric terms, providing a complete and understandable solution within the original context of the problem.