Problem 21 - Calculus Single Variable

Question

Use the method of substitution to calculate the indefinite integrals. $$ \int \frac{\sin (\sqrt{x})}{\sqrt{x}} d x $$

Step-by-Step Solution

Answer

The integral is \( -2 \cos(\sqrt{x}) + C \).

Step 1: Determine the Substitution

Notice that the integrand has a form that suggests substitution: Let's use substitution where \( u = \sqrt{x} \). This implies that \( x = u^2 \).

Step 2: Differentiate the Substitution

Differentiate the substitution to express \( dx \) in terms of \( du \): \[ \frac{d}{dx}(u) = \frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}}. \]Thus, \( du = \frac{1}{2\sqrt{x}} dx \), hence \( dx = 2u \, du \) because \( u = \sqrt{x} \).

Step 3: Substitute and Simplify the Integral

Substitute \( u = \sqrt{x} \) and \( dx = 2u \ du \) into the original integral:\[ \int \frac{\sin(\sqrt{x})}{\sqrt{x}} dx = \int \frac{\sin u}{u} \cdot 2u \, du = \int 2 \sin u \, du. \]

Step 4: Integrate with Respect to 'u'

The integral becomes:\[ 2 \int \sin u \, du. \]The integral of \( \sin u \) is \( -\cos u \), so:\[ 2 \int \sin u \, du = -2 \cos u + C, \] where \( C \) is the constant of integration.

Step 5: Substitute Back to Original Variable

Recall that \( u = \sqrt{x} \). Substitute back to express the integral in terms of \( x \):\[ -2 \cos u + C = -2 \cos(\sqrt{x}) + C. \]

Key Concepts

Indefinite Integrals

An indefinite integral, also known as an antiderivative, represents a family of functions whose derivative equals the original integrand. When you see the integral sign \( \int \), it indicates that you're trying to reverse the process of differentiation. This means for a function \( f(x) \), we look for its antiderivative \( F(x) \) such that \( F'(x) = f(x) \). Indefinite integrals are always accompanied by a constant of integration \( C \), acknowledging that there are infinitely many potential solutions. This constant arises because the derivative of a constant is zero, which means any added constant doesn't change the differentiation result.
For example, if we have \( \int 2x \, dx \), the antiderivative would be \( x^2 + C \), because \( \frac{d}{dx}(x^2 + C) = 2x \). In the context of our problem, finding \( \int \frac{\sin(\sqrt{x})}{\sqrt{x}} \, dx \) requires identifying a function whose derivative returns the given integrand.

Trigonometric Integration

Trigonometric integration deals with finding the integrals of trigonometric functions. These integrals often involve basic trigonometric identities or require specific techniques that simplify the integration process.
Common trigonometric integrals include functions like \( \int \sin(x) \, dx \) and \( \int \cos(x) \, dx \). The integral of \( \sin(x) \) is \( -\cos(x) + C \), while the integral of \( \cos(x) \) is \( \sin(x) + C \).
In the problem, after substitution, the integral included \( 2 \int \sin(u) \, du \). The integral of \( \sin(u) \) is straightforward: \( -\cos(u) \). The solution then proceeds by multiplying by the factor 2 introduced in the substitution process, leading to the final form of the integral in terms of \( u \). This involves reversing the substitution to express the result back in terms of \( x \).

• Understand common trigonometric identities to simplify problematic integrals.
• Apply standard integration rules for \( \sin \) and \( \cos \).
• Remember to include the constant of integration.

U-Substitution

U-substitution is a technique for simplifying an integral by making a substitution that transforms the problem into a more recognizable form. It is especially handy when the integral contains a composite function. The method involves:

• Identifying a part of the integrand as a new variable \( u \). This step often makes the function easy to work with.
• Finding \( du \), the derivative of \( u \), to replace \( dx \) accordingly.
• Rewriting the integral entirely in terms of \( u \).
• Integrating in terms of \( u \), then substituting back to the original variable.

In our exercise, \( u = \sqrt{x} \) led us to \( x = u^2 \), and differentiating this relationship gave us \( dx = 2u \, du \). The substitution transformed the original problem into \( \int 2 \sin(u) \, du \), which is simpler to integrate because it deals with a basic trigonometric form. Remember that after finding the integral in terms of \( u \), it's essential to substitute back \( u = \sqrt{x} \) to express the solution in terms of the original variable \( x \). This approach not only simplifies challenging integrals but also reinforces the connection between variables in calculus.