Problem 31 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 15-34, use the method of substitution to find each of the following indefinite integrals. \(\int x \cos \left(x^{2}+4\right) \sqrt{\sin \left(x^{2}+4\right)} d x\)

Step-by-Step Solution

Answer

\( \frac{1}{3} \sin(x^2 + 4)^{3/2} + C \).

Step 1: Identify the Inner Function for Substitution

We start by looking for a part of the integrand that can be substituted to simplify the integral. Here, the expression inside the trigonometric and square root functions, \( x^2 + 4 \), is a good candidate for substitution. So, let's set \( u = x^2 + 4 \).

Step 2: Differentiate the Inner Function

Next, we need to differentiate \( u = x^2 + 4 \) with respect to \( x \). This gives us \( \frac{du}{dx} = 2x \). Solving for \( dx \), we have \( dx = \frac{du}{2x} \).

Step 3: Substitute into the Integral

Replace \( x^2 + 4 \) with \( u \) and \( dx \) with \( \frac{du}{2x} \) in the integral. The integral becomes: \( \int x \cos(u) \sqrt{\sin(u)} \cdot \frac{du}{2x} \). The \( x \) terms cancel out, simplifying to \( \frac{1}{2} \int \cos(u) \sqrt{\sin(u)} \ du \).

Step 4: Integrate in Terms of u

Now integrate \( \frac{1}{2} \int \cos(u) \sqrt{\sin(u)} \ du \). To solve this, let \( v = \sin(u) \), then \( \frac{dv}{du} = \cos(u) \), which implies \( dv = \cos(u) \ du \). The integral becomes \( \frac{1}{2} \int \sqrt{v} \ dv \).

Step 5: Solve the Integral

The integral \( \int \sqrt{v} \ dv = \int v^{1/2} \ dv \) can be solved using the power rule for integration. It results in \( \frac{1}{2} \cdot \frac{2}{3} v^{3/2} = \frac{1}{3} v^{3/2} \).

Step 6: Substitute Back to Original Variable

Replace \( v \) with \( \sin(u) \) to return to the original variable \( u \). The expression becomes \( \frac{1}{3} \sin(u)^{3/2} \). Then, replace \( u \) with \( x^2 + 4 \), yielding \( \frac{1}{3} \sin(x^2 + 4)^{3/2} \).

Step 7: Add the Constant of Integration

Finally, add the constant of integration \( C \) to the solution: \( \frac{1}{3} \sin(x^2 + 4)^{3/2} + C \).

Key Concepts

method of substitution

The method of substitution is a powerful technique in calculus used to simplify the process of finding indefinite integrals. This technique resembles solving puzzles or finding a key to unlock a complex problem. It is often applied in situations where there is a recognizable part of the integrand, which can be replaced with a single variable to reduce the complexity of the integral.

When using substitution, you typically look for an "inner function" within a composite function, which is why it is sometimes referred to as the "u-substitution." The goal is to replace a complex expression with a simple variable, usually labeled as "u."

• Identify the inner function: Look for an expression inside parentheses, or under a root, exponent, or trigonometric function. This is your potential substitution "u."
• Differentiation: Differentiate the chosen inner function to find the relationship between dx and du.
• Replace and Simplify: Substitute the identified parts and integrate with respect to the new variable.
• Reverse Substitution: Once the integration is complete, replace the variable back to the original terms used in the integral.

This method is particularly useful for trigonometric, exponential, and logarithmic functions. In the given problem, choosing the inner function as the term inside the trigonometric and square root functions (\( x^2 + 4 \)), allowed for a simplified integration process.

integration techniques

Integration techniques are various methods used to calculate the integral of a function. Indefinite integrals, in particular, allow us to understand how functions accumulate values. The fundamental goal is to find the antiderivative, which is a function whose derivative matches the given integrand.

Several techniques can be employed, depending on the form of the integrand:

• Basic Antiderivatives: Recognizing the integral from standard formulas (e.g., power rule: \( \int x^n dx = \frac{x^{n+1}}{n+1} + C \)).
• Substitution: As described earlier, using substitution simplifies integrals by replacing complex expressions with simpler variables.
• Integration by Parts: Useful for products of functions, derived from the product rule for differentiation (not used in the given problem).
• Partial Fraction Decomposition: Applied to rational functions by breaking them down into simpler fractions.

In our case, substitution was the chosen technique to simplify the integral into a more manageable form, effectively reducing it to the integral of a simpler function, then solving with basic antiderivatives.

trigonometric integrals

Trigonometric integrals involve the integration of functions that include trigonometric expressions like sine, cosine, tangent, etc. These integrals often require particular strategies and techniques to solve.

For instance, if an integral contains both sine and cosine, you might consider using trigonometric identities to simplify. However, trigonometric substitution or transformation is often needed when the expression involves products or powers of trigonometric functions.

Key considerations:

• Trigonometric Identities: These can be used to simplify or transform trigonometric functions into a form that is easier to integrate.
• Substitution: Essential for simplifying expressions before integrating, as showcased in transforming the provided integral into a manageable form.
• Complex Trigonometric Expressions: Approaches like substitution seem particularly effective when faced with complex compositions, like a combination of trigonometric functions and other operations (e.g., square roots).

In the example from the original exercise, handling \( \cos \) and \( \sqrt{\sin} \) through substitution simplified the problem. By transforming the variable, the cumbersome trigonometric terms became straightforward to integrate.