Problem 6 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-12, evaluate the given integral. $$ \int \sin ^{3} x \cos x d x $$

Step-by-Step Solution

Answer

\( \int \sin^3 x \cos x \, dx = \frac{\sin^4 x}{4} + C \).

Step 1: Identify the Substitution

Given the integral \( \int \sin^3 x \cos x \, dx \), we notice that the integral has a function \( \sin x \) raised to a power and its derivative (\( \cos x \)) present. This suggests using substitution. We choose \( u = \sin x \), which implies \( du = \cos x \, dx \).

Step 2: Substitution

Substitute \( u = \sin x \) and \( du = \cos x \, dx \) into the integral. The integral becomes \( \int u^3 \, du \).

Step 3: Integrate the New Expression

Integrate \( \int u^3 \, du \). This is a standard power rule integration: \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \). Thus, \( \int u^3 \, du = \frac{u^4}{4} + C \).

Step 4: Substitute Back to Original Variable

Replace \( u \) with \( \sin x \) to return to the original variable. Therefore, the solution is \( \frac{\sin^4 x}{4} + C \).

Key Concepts

Substitution Method

The substitution method is a powerful technique in calculus integration. It is used to simplify the integration process by changing variables. When we encounter an integral that is complex, substitution helps in translating it into a more manageable form.

Here's what happens step-by-step:

• Identify part of the integrand that can be substituted. Usually, this is a function and its derivative, for example, if you see a function like \( f(g(x))g'(x) \), consider \( u = g(x) \).
• Find the differential \( du \) which is the derivative of \( u \, du = g'(x)dx\).
• Rewrite the entire integral in terms of \( u \), consequently transforming the problem.
• Perform the integration with respect to \( u \).
• Substitute back the original variable to finish solving the integral.

In our example, choosing \( u = \sin x \) and recognizing \( \cos x \, dx = du \), made the integration straightforward. It converted \( \int \sin^3 x \, \cos x \, dx \) into the simpler \( \int u^3 \, du \). This simplification using substitution is what makes complex integrals more approachable.

Trigonometric Integrals

Trigonometric integrals involve the integration of products of trigonometric functions. They can often be intricate due to their cyclical nature. However, patterns and relationships between the trigonometric functions can be utilized to simplify the integration.

In our example of \( \int \sin^3 x \, \cos x \, dx \), we dealt with a product of sine and cosine. Understanding the relations between these functions and their derivatives is crucial. Here are some useful tips when working with trigonometric integrals:

• Separate the powers if possible. For example, \( \sin^3 x = \sin^2 x \, \sin x \).
• Use trigonometric identities to simplify expressions. For instance, \( \sin^2 x = 1 - \cos^2 x \).
• When recognizing that one of the trigonometric terms is the derivative of the other (e.g., \( \cos x \) is the derivative of \( \sin x \)), consider the substitution method as used in our example.

The key is to often convert the trigonometric terms into simpler expressions or to take advantage of the presence of a derivative, as was accomplished by substituting \( \sin x \) in our problem.

Power Rule Integration

Power rule integration is a fundamental technique used for integrating functions of the form \( x^n \). It's a direct way to solve integrals, making it an essential tool in the calculus toolbox.

The power rule states:

• If \( n \) is any real number except -1, then \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \).
• This rule simplifies the integration process by increasing the exponent by one and dividing by the new exponent.

In the step-by-step solution presented, after substitution, we transformed the integral into \( \int u^3 \, du \). Applying the power rule here was straightforward: since the form matched \( u^n \, \frac{u^{4}}{4}\ + C \).

Always remember, after applying the power rule, to revert to the original variable when using substitution. This brings the solution back to the terms of the original integral, completing the integration process.